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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Miscellaneous Exercise 3 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Miscellaneous Exercise 3 Questions And Answers Maharashtra Board

I) Select the correct option from the given alternatives.
Question 1.
The principal of solutions equation sinθ = \(\frac{-1}{2}\) are ________.

Solution:
(b) \(\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

Question 2.
The principal solution of equation cot θ = \(\sqrt {3}\) ___________.

Solution:
(a) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)

Question 3.
The general solution of sec x = \(\sqrt {2}\) is __________.
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z
(b) 2nπ ± \(\frac{\pi}{2}\), n ∈ Z
(c) nπ ± \(\frac{\pi}{2}\), n ∈ Z
(d) 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Solution:
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z

Question 4.
If cos pθ = cosqθ, p ≠ q rhen ________.
(a) θ = \(\frac{2 n \pi}{p \pm q}\)
(b) θ = 2nπ
(c) θ = 2nπ ± p
(d) nπ ± q
Solution:
(a) θ = \(\frac{2 n \pi}{p \pm q}\)

Question 5.
If polar co-ordinates of a point are \(\left(2, \frac{\pi}{4}\right)\) then its cartesian co-ordinates are ______.
(a) (2, \(\sqrt {2}\) )
(b) (\(\sqrt {2}\), 2)
(c) (2, 2)
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))
Solution:
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))

Question 6.
If \(\sqrt {3}\) cosx – sin x = 1, then general value of x is _________.
(a) 2nπ ± \(\frac{\pi}{3}\)
(b) 2nπ ± \(\frac{\pi}{6}\)
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)
(d) nπ + (-1)ⁿ\(\frac{\pi}{3}\)
Solution:
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)

Question 7.
In ∆ABC if ∠A = 45°, ∠B = 60° then the ratio of its sides are _________.
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
(b) \(\frac{\pi}{2}\) : 2 : \(\frac{\pi}{3}\) + 1
(c) 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\)
(d) 2 : 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
Solution:
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1

Question 8.
In ∆ABC, if c² + a² – b² = ac, then ∠B = __________.
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(b) \(\frac{\pi}{3}\)

Question 9.
In ABC, ac cos B – bc cos A = ____________.
(a) a² – b²
(b) b² – c²
(c) c² – a²
(d) a² – b² – c²
Solution:
(a) a² – b²

Question 10.
If in a triangle, the are in A.P. and b : c = \(\sqrt {3}\) : \(\sqrt {2}\) then A is equal to __________.
(a) 30°
(b) 60°
(c) 75°
(d) 45°
Solution:
(c) 75°

Question 11.
cos^-1\(\left(\cos \frac{7 \pi}{6}\right)\) = ________.

Question 12.
The value of cot (tan^-1 2x + cot^-1 2x) is __________.
(a) 0
(b) 2x
(c) π + 2x
(d) π – 2x
Solution:
(a) 0

Question 13.
The principal value of sin^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) is ____________.

Solution:
(d) \(-\frac{\pi}{3}\)

Question 14.
If sin^-1\(\frac{4}{5}\) + cos^-1\(\frac{,12}{13}\) = sin^-1 ∝, then ∝ = _____________.
(a) \(\frac{63}{65}\)
(b) \(\frac{62}{65}\)
(c) \(\frac{61}{65}\)
(d) \(\frac{60}{65}\)
Solution:
(a) \(\frac{63}{65}\)

Question 15.
If tan^-1(2x) + tan^-1(3x) = \(\frac{\pi}{4}\), then x = ________.
(a) -1
(b) \(\frac{1}{6}\)
(c) \(\frac{2}{6}\)
(d) \(\frac{3}{2}\)
Solution:
(b) \(\frac{1}{6}\)

Question 16.
2 tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{7}\) = ______.
(a) tan^-1\(\frac{4}{5}\)
(b) \(\frac{\pi}{2}\)
(c) 1
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

Question 17.
tan (2 tan^-1\(\left(\frac{1}{5}\right)-\frac{\pi}{4}\)) = ______.
(a) \(\frac{17}{7}\)
(b) \(-\frac{17}{7}\)
(c) \(\frac{7}{17}\)
(d) \(-\frac{7}{17}\)
Solution:
(d) \(-\frac{7}{17}\)

Question 18.
The principal value branch of sec^-1 x is __________.

Solution:
(b) [0, π] – {\(\frac{\pi}{2}\)}

Question 19.
cos[tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{2}\)] = ________.
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{\pi}{4}\)
Solution:
(a) \(\frac{1}{\sqrt{2}}\)

Question 20.
If tan θ + tan 2θ + tan 3θ = tan θ∙tan 2θ∙tan 3θ, then the general value of the θ is _______.
(a) nπ
(b) \(\frac{n \pi}{6}\)
(c) nπ ± \(\frac{n \pi}{4}\)
(d) \(\frac{n \pi}{2}\)
Solution:
(b) \(\frac{n \pi}{6}\)
[Hint: tan(A + B + C) = \(\frac{\tan A+\tan B+\tan C-\tan A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}\)
Since , tan θ + tan 2θ + tan 3θ = tan θ ∙ tan 2θ ∙ tan 3θ,
we get, tan (θ + 2θ + 3θ) = θ
∴ tan6θ = 0
∴ 6θ = nπ, θ = \(\frac{n \pi}{6}\).]

Question 21.
If any ∆ABC, if a cos B = b cos A, then the triangle is ________.
(a) Equilateral triangle
(b) Isosceles triangle
(c) Scalene
(d) Right angled
Solution:
(b) Isosceles triangle

II: Solve the following
Question 1.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{2}\)
Solution:
sin2θ = \(-\frac{1}{2}\)
Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

(ii) tan3θ = -1
Solution:
Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)
= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]
∴ tan3θ = tan\(\frac{3 \pi}{4}\) = tan\(\frac{7 \pi}{4}\) = tan\(\frac{11 \pi}{4}\) = tan\(\frac{15 \pi}{4}\)

(iii) cotθ = 0
Solution:
cotθ = 0
Since θ ∈ (0, 2π),
cotθ = 0 = cot \(\frac{\pi}{2}\) = cot (π + \(\frac{\pi}{2}\) …[∵ cos(π + θ) = cotθ]
∴ cotθ = cot\(\frac{\pi}{2}\) = cot\(\frac{3 \pi}{2}\)
∴ θ = \(\frac{\pi}{2}\) or θ = \(\frac{3 \pi}{2}\)
Hence, the required principal solutions are \(\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}\)

Question 2.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{\sqrt{2}}\)
Solution:

(ii) tan5θ = -1
Solution:

(iii) cot2θ = 0
Solution:

Question 3.
Which of the following equations have no solutions ?
(i) cos 2θ = \(\frac{1}{3}\)
Solution:
cos 2θ = \(\frac{1}{3}\)
Since \(\frac{1}{3}\) ≤ cosθ ≤ 1 for any θ
cos2θ = \(\frac{1}{3}\) has solution

(ii) cos² θ = -1
Solution:
cos2θ = -1
This is not possible because cos2θ ≥ 0 for any θ.
∴ cos2θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 sin θ = 5
Solution:
3 sin θ = 5
∴ sin θ = \(\frac{5}{3}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

Question 4.
Find the general solutions of the following equations :
(i) tanθ = \(-\sqrt {x}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z.
Now, tanθ = \(-\sqrt {x}\)
∴ tanθ = tan\(\frac{\pi}{3}\) …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tanθ = tan\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ tan(π – θ) = -tanθ]
∴ tanθ = tan\(\frac{2 \pi}{3}\)
∴ the required general solution is
θ = nπ + \(\frac{2 \pi}{3}\), n ∈ Z.

(ii) tan²θ = 3
Solution:
The general solution of tan²θ = tan²∝ is
θ = nπ ± ∝, n ∈ Z.
Now, tan²θ = 3 = (\(\sqrt {x}\))²
∴ tan²θ = (tan\(\frac{\pi}{3}\))² …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tan²θ = tan²\(\frac{\pi}{3}\)
∴ the required general solution is
θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) sin θ – cosθ = 1
Solution:
∴ cosθ – sin θ = -1

(iv) sin²θ – cos²θ = 1
Solution:
sin²θ – cos²θ = 1
∴ cos²θ – sin²θ = -1
∴ cos2θ = cosπ …(1)
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of (1) is given by
2θ = 2nπ ± π, n ∈ Z
∴ θ = nπ ± \(\frac{\pi}{2}\), n ∈ Z

Question 5.
In ∆ABC prove that cos \(\left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)\) sin \(\frac{C}{2}\)
Solution:
By the sine rule,

Question 6.
With usual notations prove that \(\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{c^{2}}\).
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC

Question 7.
In ∆ABC prove that (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\) = c².
Solution:
LHS (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b² – 2ab) cos²\(\frac{\mathrm{C}}{2}\) + (a² + b² + 2ab) sin\(\frac{\mathrm{C}}{2}\)²
= (a² + b²) cos²\(\frac{\mathrm{C}}{2}\) – 2ab cos²\(\frac{\mathrm{C}}{2}\) + (a² + b²) sin²\(\frac{\mathrm{C}}{2}\) + 2ab sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b²) (cos²\(\frac{\mathrm{C}}{2}\) + sin²\(\frac{\mathrm{C}}{2}\)) – 2ab(cos²\(\frac{\mathrm{C}}{2}\) – sin²\(\frac{\mathrm{C}}{2}\))
= a² + b² – 2ab cos C
= c² = RHS.

Question 8.
In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.
Solution:
cos A= sin B – cos C
∴ cos A + cos C = sin B

∴ A – C = B
∴ A = B + C
∴ A + B + C = 180° gives
A + A = 180°
∴ 2A = 180 ∴ A = 90°
∴ ∆ ABC is a rightangled triangle.

Question 9.
If \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\) then show that a², b², c², are in A.P.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb,sin C = kc
Now, \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
∴ sinA∙sin(B – C) = sinC∙sin(A -B)
∴ sin [π – (B + C)] ∙ sin (B – C)
= sin [π – (A + B)]∙sin (A – B) … [∵ A + B + C = π]
∴ sin(B + C) ∙ sin(B – C) = sin (A + B) ∙ sin (A – B)
∴ sin²B – sin²C = sin²A – sin²B
∴ 2 sin²B = sin²A + sin²C
∴ 2k²b² = k²a² + k²c²
∴ 2b² = a² + c²
Hence, a², b², c² are in A.P.

Question 10.
Solve the triangle in which a = (\(\sqrt {3}\) + 1), b = (\(\sqrt {3}\) – 1) and ∠C = 60°.
Solution:
Given : a = \(\sqrt {3}\) + 1, b = \(\sqrt {3}\) – 1 and ∠C = 60°.
By cosine rule,
c² = a² + b² – 2ab cos C
= (\(\sqrt {3}\) + 1)² + (\(\sqrt {3}\) – 1)² – 2(\(\sqrt {3}\) + 1)(\(\sqrt {3}\) – 1)cos60°
= 3 + 1 + 2\(\sqrt {3}\) + 3+ 1 – 2\(\sqrt {3}\) – 2(3 – 1)\(\left(\frac{1}{2}\right)\)
= 8 – 2 = 6
∴ c = \(\sqrt {6}\) …[∵ c > 0)
By sine rule,

∴ sin A = sin 60° cos 45° + cos 60° sin 45°
and sin B = sin 60° cos 45° – cos 60° sin 45°
∴ sin A = sin (60° + 45°) – sin 105°
and sin B = sin (60° – 45°) = sin 15°
∴ A = 105° and B = 15°
Hence, A = 105°, B 15° and C = \(\sqrt {6}\) units.

Question 11.
In ∆ABC prove the following :
(i) a sin A – b sin B = c sin (A – B)
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC,
LHS = a sin A – b sinB
= ksinA∙sinA – ksinB∙sinB
= k (sin²A – sin²B)
= k (sin A + sin B)(sin A – sin B)

= k × sin (A + B) × sin (A – B)
= ksin(π – C)∙sin(A – B) … [∵ A + B + C = π]
= k sinC∙sin (A – B)
= c sin (A – B) = RHS.

(ii) \(\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}\).
Solution:

(iii) a² sin (B – C) = (b² – c²) sinA
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC
RHS = (b² – c²) sin A
= (k²sin²B – k²sin²C)sin A
= k²(sin²B – sin²C) sin A
= k²(sin B + sin C)(sin B – sin C) sin A

= k² × sin (B + C) × sin (B – C) × sin A
= k²∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]
= k²sin A∙sin (B – C)∙sin A
= (k sin A)²∙sin (B – C)
= a²sin (B – C) = LHS.

(iv) ac cos B – bc cos A = (a² – b²).
Solution:
LHS = ac cos B – bc cos A
= ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) – bc\(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\)
=\(\frac{1}{2}\)(c² + a² – b²) – \(\frac{1}{2}\)(b² + c² – a²)
= \(\frac{1}{2}\)(c² + a² – b² – b² – c² + a²)
= \(\frac{1}{2}\)(2a² – 2b²) = a² – b2 = RHS.

(v) \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\) .
Solution:

(vi) \(\frac{\cos 2 \mathrm{~A}}{a^{2}}-\frac{\cos 2 \mathrm{~B}}{b^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}\).
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}\)

(vii) \(\frac{b-c}{a}=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}\)
Solution:
By sine rule,

Question 12.
In ∆ABC if a², b², c², are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Question is modified
In ∆ABC if a, b, c, are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Solution:
a, b, c, are in A.P.
∴ 2b = a + c …(1)

Question 13.
In ∆ABC if ∠C = 90º then prove that sin(A – B) = \(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)
Solution:
In ∆ABC, if ∠C = 90º
∴ c² = a² + b² …(1)
By sine rule,

Question 14.
In ∆ABC if \(\frac{\cos A}{a}=\frac{\cos B}{b}\), then show that it is an isosceles triangle.
Solution:
Given : \(\frac{\cos A}{a}=\frac{\cos B}{b}\) ….(1)
By sine rule,

∴ sin A cos B = cos A sinB
∴ sinA cosB – cosA sinB = 0
∴ sin (A – B) = 0 = sin0
∴ A – B = 0 ∴ A = B
∴ the triangle is an isosceles triangle.

Question 15.
In ∆ABC if sin²A + sin²B = sin²C then prove that the triangle is a right angled triangle.
Question is modified
In ∆ABC if sin²A + sin²B = sin²C then show that the triangle is a right angled triangle.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sinB = kb, sin C = kc
∴ sin²A + sin²B = sin²C
∴ k²a² + k²b² = k²c²
∴ a² + b² = c²
∴ ∆ABC is a rightangled triangle, rightangled at C.

Question 16.
In ∆ABC prove that a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0.
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
LHS = a²(cos²B – cos²C) + b²( cos²C – cos²A) + c²(cos²A – cos²B)
= k²sin²A [(1 – sin²B) – (1 – sin²C)] + k²sin²B [(1 – sin²C) – (1 – sin²A)] + k²sin²C[(1 – sin²A) – (1 – sin²B)]
= k²sin²A (sin²C – sin²B) + k²sin²B(sin²A – sin²C) + k²sin²C (sin²B – sin²A)
= k²(sin²A sin²C – sin²Asin²B + sin²A sin²B – sin²B sin²C + sin²B sin²C – sin²A sin²C)
= k²(0) = 0 = RHS.

Question 17.
With usual notations show that (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C.
Solution:
By sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = fksinA, b = ksinB, c = ksinC


From (1), (2) and (3), we get
(c² – a² + b²) tan A = (a² – b² + c²) tan B
= (b² – c² + a²) tan C.

Question 18.
In ∆ABC, if a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\), then prove that a , b ,c are in A.P.
Solution:
a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\)

∴ a + c + b = 3b …[∵ a cos C + c cos A = b]
∴ a + c = 2b
Hence, a, b, c are in A.P.

Question 19.
Show that 2 sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{24}{7}\right)\).
Solution:
Let sin²\(\left(\frac{3}{5}\right)\) = x.



∴ tan^-1\(\left(\frac{24}{7}\right)\) = RHS

Question 20.
Show that tan^-1\(\left(\frac{1}{5}\right)\) + tan^-1\(\left(\frac{1}{7}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) + tan^-1\(\left(\frac{1}{8}\right)\) = \(\frac{\pi}{4}\).
Solution:

Question 21.
Prove that tan^-1\(\sqrt {x}\) = \(\frac{1}{2}\) cos^-1\(\left(\frac{1-x}{1+x}\right)\), if x ∈ [0, 1].
Solution:
Let tan^-1\(\sqrt {x}\) = y
∴ tan y = \(\sqrt {x}\) ∴ x = tan²y

Question 22.
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\frac{1}{3}\) = \(\frac{9}{4}\) sin^-1\(\frac{2 \sqrt{2}}{3}\).
Question is modified
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\left(\frac{1}{3}\right)\) = \(\frac{9}{4}\) sin^-1\(\left(\frac{2 \sqrt{2}}{3}\right)\).
Solution:
We have to show that

Question 23.
Show that

Solution:

Question 24.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Solution:
sin(sin^-1\(\frac{1}{5}\) = 1

Question 25.
If tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\) then find the value of x.
Solution:
tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\)

∴ x = ±\(\frac{1}{\sqrt{2}}\).

Question 26.
If 2 tan^-1(cos x ) = tan^-1(cosec x) then find the value of x.
Solution:
2 tan^-1(cos x ) = tan^-1(cosec x)

Question 27.
Solve: tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x), for x > 0.
Solution:
tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x)

Question 28.
If sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\)

Question 29.
If tan^-12x + tan^-13x = \(\frac{\pi}{4}\), then find the value of x.
Question is modified
If tan^-12x + tan^-13x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
tan^-12x + tan^-13x = \(\frac{\pi}{4}\)
∴ tan^-1\(\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)\) = tan\(\frac{\pi}{4}\), where 2x > 0, 3x > 0
∴ \(\frac{5 x}{1-6 x^{2}}\) = tan\(\frac{\pi}{4}\) = 1
∴ 5x = 1 – 6x²
∴ 6x² + 5x – 1 = 0
∴ 6x² + 6x – x – 1 = 0
∴ 6x(x +1) – 1(x + 1) = 0
∴ (x + 1)(6x – 1) = 0
∴ x = -1 or x = \(\frac{1}{6}\)
But x > 0 ∴ x ≠ -1
Hence, x = \(\frac{1}{6}\)

Question 30.
Show that tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\) = tan^-1\(\frac{2}{9}\).
Solution:
LHS = tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\)

Question 31.
Show that cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\) = cot^-1\(\frac{3}{4}\).
Solution:
LHS = cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\)

Question 32.
Show that tan^-1\(\frac{1}{2}\) = \(\frac{1}{3}\) tan^-1\(\frac{11}{2}\).
Solution:
We have to show that

Question 33.
Show that cos^-1\(\frac{\sqrt{3}}{2}\) + 2sin^-1\(\frac{\sqrt{3}}{2}\) = \(\frac{5 \pi}{6}\)
Solution:

Question 34.
Show that 2cot^-1\(\frac{3}{2}\) + sec^-1\(\frac{13}{12}\) = \(\frac{\pi}{2}\)
Solution:

Question 35.
Prove the following :
(i) cos^-1 x = tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Question is modified
cos^-1 x = tan^-1\(\left(\frac{\sqrt{1-x^{2}}}{x}\right)\), if x > 0.
Solution:

(ii) cos^-1 x = π + tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Solution:

Question 36.
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\frac{2 x}{1-x^{2}}\) = sin^-1\(\frac{2 x}{1+x^{2}}\) = cos^-1\(\frac{1-x^{2}}{1+x^{2}}\)
Question is modified
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\left(\frac{2 x}{1-x^{2}}\right)\) = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\) = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let tan^-1x = y
Then, x = tany

Question 37.
If x, y, z, are positive then prove that tan^-1\(\frac{x-y}{1+x y}\) + tan^-1\(\frac{y-z}{1+y z}\) + tan^-1\(\frac{z-x}{1+z x}\) = 0
Solution:

Question 38.
If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\) then, show that xy + yz + zx = 1
Solution:
tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\)

∴ 1 – xy – yz – zx = 0
∴ xy + yz + zx = 1.

Question 39.
If cos^-1 x + cos^-1 y + cos^-1 z = π then show that x² + y² + z² + 2xyz = 1.
Solution:
0 ≤ cos^-1x ≤ π and
cos^-1x + cos^-1y+ cos^-1z = 3π
∴ cos^-1x = π, cos^-1y = π and cos^-1z = π
∴ x = y = z = cosπ = -1
∴ x² + y² + z² + 2xyz
= (-1)² + (-1)² + (-1)² + 2(-1)(-1)(-1)
= 1 + 1 + 1 – 2
= 3 – 2 = 1.

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Trigonometric Functions Ex 3.1 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.2 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.3 Class 12 Maths Solutions
• Trigonometric Functions Miscellaneous Exercise 3 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.1 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.2 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.3 Class 12 Maths Solutions
• Pair of Straight Lines Miscellaneous Exercise 4 Class 12 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.3 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Exercise 3.3 Questions And Answers Maharashtra Board

Question 1.
Find the principal values of the following :
(i) sin^-1\(\left(\frac{1}{2}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1\(\left(\frac{1}{2}\right)\) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\)
∴ sin∝ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ – \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1\(\left(\frac{1}{2}\right)\) is \(\frac{\pi}{6}\).

(ii) cosec^-1(2)
Solution:
The principal value branch of cosec^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec^-1(2) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\), ∝ ≠ 0
∴ cosec^-1 ∝ = 2 = cosec\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of cosec^-1(2) is \(\frac{\pi}{6}\).

(iii) tan^-1(-1)
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Let tan^-1(-1) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = -1 = -tan\(\frac{\pi}{4}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{4}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{4}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-1) is –\(\frac{\pi}{4}\).

(iv) tan^-1(-\(\sqrt {3}\))
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Let tan^-1(-\(\sqrt {3}\)) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = –\(\sqrt {3}\) = -tan\(\frac{\pi}{3}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{3}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{3}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{3}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-\(\sqrt {3}\)) is –\(\frac{\pi}{3}\).

(v) sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ sin∝ = \(\left(\frac{1}{\sqrt{2}}\right)\) = sin\(\frac{\pi}{4}\)
∴ ∝ = \(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

(vi) cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:
The principal value branch of cos^-1x is (0, π).
Let cos^-1\(\left(-\frac{1}{2}\right)\) = ∝, where 0 ≤ ∝ ≤ π
∴ cos∝ = \(-\frac{1}{2}\) = -cos\(\frac{\pi}{3}\)
∴ cos∝ = cos\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ cos(π – θ) = -cosθ)
∴ cos∝ = cos\(\frac{2 \pi}{3}\)
∴ ∝ = \(\frac{2 \pi}{3}\) …[∵ 0 ≤ \(\frac{2 \pi}{3}\) ≤ π]
∴ the principal value of cos^-1\(\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).

Question 2.
Evaluate the following :
(i) tan^-1(1) + cos^-1\(\left(\frac{1}{2}\right)\) + sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(ii) cos^-1\(\left(\frac{1}{2}\right)\) + 2 sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(iii) tan^-1\(\sqrt {3}\) – sec^-1(-2)
Solution:

∴ tan^-1\(\sqrt {3}\) – sec^-1(-2)
= \(\frac{\pi}{3}-\frac{2 \pi}{3}\) …[By (1) and (2)]
= –\(\frac{\pi}{3}\).

(iv) cosec^-1( \(-\sqrt{2}\)) + cot^-1(\(\sqrt{3}\))
Solution:

Question 3.
Prove the following :
(i) sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(-\frac{3 \pi}{4}\)
Question is modified.
sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(\frac{3 \pi}{4}\)
Solution:

(ii) sin^-1\(\left(-\frac{1}{2}\right)\) + cos^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) = cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:

(iii) sin^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{12}{13}\right)\) = sin^-1\(\left(\frac{56}{65}\right)\)
Solution:

(iv) cos^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{4}{5}\right)\) = \(\frac{\pi}{2}\)
Solution:
Let cos^-1\(\left(\frac{3}{5}\right)\) = x
∴ cosx = \(\left(\frac{3}{5}\right)\), where 0 < x < \(\frac{\pi}{2}\) ∴ sinx > 0

(v) tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\)
Solution:
LHS = tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\)
= tan^-1\(\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\)
= tan^-1\(\left(\frac{3+2}{6-1}\right)\) = tan^-1(1)
= tan^-1\(\left(\tan \frac{\pi}{4}\right)\) = \(\frac{\pi}{4}\)
= RHS.

(vi) 2 tan^-1\(\left(\frac{1}{3}\right)\) = tan^-1\(\left(\frac{3}{4}\right)\)
Solution:

(vii) tan^-1\(\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]\) = \(\frac{\pi}{4}\) + θ if θ ∈ \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)
Solution:

(viii) tan^-1\(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\theta}{2}\), if θ ∈ (0, π)
Solution:

= \(\frac{\theta}{2}\) …[∵ tan^-1(tanθ) = θ]
= RHS.

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Trigonometric Functions Ex 3.1 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.2 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.3 Class 12 Maths Solutions
• Trigonometric Functions Miscellaneous Exercise 3 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.1 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.2 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.3 Class 12 Maths Solutions
• Pair of Straight Lines Miscellaneous Exercise 4 Class 12 Maths Solutions

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Miscellaneous Exercise 2 Questions and Answers.

11th Maths Part 1 Trigonometry – I Miscellaneous Exercise 2 Questions And Answers Maharashtra Board

I. Select the correct option from the given alternatives.

Question 1.
The value of the expression
cos1°. cos2°. cos3° … cos 179° =
(A) -1
(B) 0
(C) \(\frac{1}{\sqrt{2}}\)
(D) 1
Answer:
(B) 0

Explanation:
cos 1° cos 2° cos 3° … cos 179°
= cos 1° cos 2° cos 3° … cos 90°… cos 179°
= 0 …[∵ cos 90° = 0]

Question 2.
\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to
(A) 2cosec A
(B) 2 sec A
(C) 2 sin A
(D) 2 cos A
Answer:
(A) 2cosec A

Explanation:

Question 3.
If α is a root of 25cos² θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to
(A) \(-\frac{24}{25}\)
(B) \(-\frac{13}{18}\)
(C) \(\frac{13}{18}\)
(D) \(\frac{24}{25}\)
Answer:
(A) \(-\frac{24}{25}\)

Explanation:

25 cos² θ + 5 cos θ – 12 = 0
∴ (5cos θ + 4) (5 cos θ – 3) = 0
∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π,
cos α < 0
∴ cos α = \(-\frac{4}{5}\)
sin² α = 1 – cos² α = 1 – \(\frac{16}{25}=\frac{9}{25}\)
∴ sin α = \(\pm \frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π sin α > 0
∴ sin α = 3/5
sin 2 α = 2 sin α cos α
= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.
If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to
(A) \(\frac{\sqrt{3}}{2}\)
(B) \(\frac{2}{\sqrt{3}}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt{3}\)
Answer:
(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.
If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to
(A) 2
(B) mn
(C) 2m
(D) 2n
Answer:
(A) 2
Explanation:

Question 6.
If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is
(A) \(\frac{14}{25}\)
(B) \(\frac{20}{21}\)
(C) \(\frac{21}{20}\)
(D) \(\frac{15}{16}\)
Answer:
(B) \(\frac{20}{21}\)

Explanation:
cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1
∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)
Subtracting (ii) from (i), we get
2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)
∴ cot θ = \(\frac{21}{20}\)
∴ tan θ = \(\frac{20}{21}\)

Question 7.
\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals
(A) 0
(B) 1
(C) sin θ
(D) cos θ
Answer:
(D) cos θ

Explanation:

Question 8.
If cosec θ – cot θ = q, then the value of cot θ is
(A) \(\frac{2 q}{1+q^{2}}\)
(B) \(\frac{2 q}{1-q^{2}}\)
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)
(D) \(\frac{1+q^{2}}{2 q}\)
Answer:
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ (cosec θ + cot θ)q = 1
∴ cosec θ + cot θ = 1/q …….(ii)
Subtracting (i) from (ii), we get
2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)
∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.
The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in
(A) A.P.
(B) G.P.
(C) H.P.
(D) Not in progression
Answer:
(B) G.P.

Explanation:

Question 10.
The value of tan 1°.tan 2° tan 3° equal to
(A) -1
(B) 1
(C) \(\frac{\pi}{2}\)
(D) 2
Answer:
(B) 1

Explanation:

tan1° tan2° tan3° … tan89°
= (tan 1° tan 89°) (tan 2° tan 88°)
…(tan 44° tan 46°) tan 45°
= (tan 1 ° cot 1 °) (tan 2° cot 2°)
…(tan 44° cot 44°) . tan 45°
…tan(∵ 90° – θ) = cot θ]
= 1 x 1 x 1 x … x 1 x tan 45° =1

II. Answer the following:

Question 1.
Find the trigonometric functions of:
90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°
Solution:
Angle of measure 90° :
Let m∠XOA = 90°
Its terminal arm (ray OA)
intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1
sin 90° = y = 1
cos 90° = x = 0
tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined
cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1
sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined
cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 0

Angle of measure 120° :
Let m∠XOA =120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

Angle of measure 225° :
Let m∠XOA = 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :
Let m∠XOA = 270°
Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).
x = 0 andy = – 1
sin 270° = y = -1
cos 270° = x = 0
tan 270° = \(\frac{y}{x}\)

Angle of measure 315° :
Let m∠XOA = 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1


[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :
Let m∠XOA = – 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (-180°):
Let m∠XOA = – 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).
∴ x = – 1 andy = 0
sin (-180°) = y = 0
cos (-180°) = x
= -1

Angle of measure (- 210°):
Let m∠XOA = -210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (- 300°):
Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 1st quadrant, x>0,y>0
x = OM = \(\frac{1}{2}\) and
y = PM = \(\frac{\sqrt{3}}{2}\)

Angle of measure (- 330°):
Let m∠XOA = – 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Question 2.
State the signs of:
i. cosec 520°
ii. cot 1899°
iii. sin 986°
Solution:
i. 520° =360° + 160°
∴ 520° and 160° are co-terminal angles.
Since 90° < 160° < 180°,
160° lies in the 2nd quadrant.
∴ 520° lies in the 2nd quadrant,
∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°
∴ 1899° and 99° are co-terminal angles.
Since 90° < 99° < 180°,
99° lies in the 2nd quadrant.
∴ 1899° lies in the 2nd quadrant.
∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°
∴ 986° and 266° are co-terminal angles.
Since 180° < 266° < 270°,
266° lies in the 3rd quadrant.
∴ 986° lies in the 3rd quadrant.
∴ sin 986° is negative.

Question 3.
State the quadrant in which 6 lies if
i. tan θ < 0 and sec θ > 0
ii. sin θ < 0 and cos θ < 0
iii. sin θ > 0 and tan θ < 0
Solution:
i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0
sec θ is positive in 1st and 4th quadrants.
∴ θ lies in the 4th quadrant.

ii. sin θ < 0
sin θ is negative in 3rd and 4th quadrants, cos θ < 0
cos θ is negative in 2nd and 3rd quadrants.
.’. θ lies in the 3rd quadrant.

iii. sin θ > 0
sin θ is positive in 1st and 2nd quadrants, tan θ < 0
tan θ is negative in 2nd and 4th quadrants.
∴ θ lies in the 2nd quadrant.

Question 4.
Which is greater?
sin (1856°) or sin (2006°)
Solution:
1856° = 5 x 360° + 56°
∴ 1856° and 56° are co-terminal angles.
Since 0° < 56° < 90°, 56° lies in the 1st quadrant.
∴ 1856° lies in the 1st quadrant,
∴ sin 1856° >0 …(i)
2006° = 5 x 360° + 206°
∴ 2006° and 206° are co-terminal angles.
Since 180° < 206° < 270°,
206° lies in the 3rd quadrant.
∴ 2006° lies in the 3rd quadrant,
∴ sin 2006° <0 …(ii)
From (i) and (ii),
sin 1856° is greater.

Question 5.
Which of the following is positive?
sin(-310°) or sin(310°)
Solution:
Since 270° <310° <360°,
310° lies in the 4th quadrant.
∴ sin (310°) < 0
-310° = -360°+ 50°
∴ 50° and – 310° are co-terminal angles.
Since 0° < 50° < 90°, 50° lies in the 1st quadrant.
∴ – 310° lies in the 1st quadrant.
∴ sin (- 310°) > 0
∴ sin (- 310°) is positive.

Question 6.
Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.
Solution:
1 – 2 sin θ cos θ
= sin² θ + cos² θ – 2sin θ cos θ
= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

Question 7.
Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R.
Solution:

Question 8.
If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.
Solution:
Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
we know that
cos²θ = 1 – sin² θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.
If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)
Solution:
Given sec θ = \(\sqrt{2}\)
We know that,
tan² θ = sec² θ – 1
= (\(\sqrt{2}\)) – 1
= 2 – 1 = 1
∴ tan θ = ±1
Since \(\frac{3 \pi}{2}\) < θ < 2π
θ lies in the 4th quadrant.
∴ tan θ < 0
∴ tan θ = -1

Question 10.
Prove the following:

i. sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1
Solution:
L.H.S. = sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A (cos² B + sin² B) + cos² A (sin² B + cos² B)
= sin²A(1) + cos²A(1)
= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)
Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)
Solution:
L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)
= (tanθ + secθ)² + (tanθ – secθ)²
= tan² θ + 2 tan θ sec θ + sec² θ
+ tan² θ – 2 tan θ sec θ +.sec² θ
= 2(tan² θ + sec² θ)

iv. 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ
Solution:
LHS.
= 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  = 2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²
= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ)
+ (1+ cot² θ)²
= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ)
– 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ
= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2
– 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ
= cot⁴ θ – tan⁴ θ = R.H.S.

v. sin⁴ θ + cos⁴ θ = sin⁴ θ + cos⁴ θ
Solution:
L.H.S. = sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ
… [ v a² + b² = (a + b)² – 2ab]
= 1 – 2sin² θ cos² θ
= R.H.S.

vi. 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0
L.H.S =
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0
= sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin2 0 + cos2 0)
…[••• a³ + b³ = (a + b)³ – 3ab(a + b)]
= (1)³ – 3 sin² θ cos² θ(1)
= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos² θ
…[Y a² + b² = (a + b)² – 2ab]
= 1-2 sin² θ cos² θ
L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1
= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c
= R.H.S.

vii. cos⁴ θ – sin⁴ θ + 1 = 2cos²θ
L.H.S. = cos⁴ θ – sin⁴ θ + 1
= (cos² θ)² – (sin² θ)² + 1 = (cos²θ + sin²θ) c(os² θ – sin²θ) +1
= (1) (cos²θ – sin²θ) + 1 = cos² θ + (1 – sin²θ)
= cos² θ + cos²θ = 2cos²θ = R.H.S.

viii. sin⁴θ + 2sin²θ cos²θ = 1 – cos⁴θ
L.H.S. = sin⁴θ + 2sin²θ cos²θ = sin²θ(sin²θ + 2cos²θ)
= (sin²θ) (sin²θ + cos²θ + cos²θ) = (1 – cos²θ) (1 + cos²θ)
= 1 – cos⁴θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)
Solution:

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)
= 2 (sin² θ + cos² θ)
= 2(1)
= 2 = R.H.S.

x. tan² θ – sin² θ = sin⁴ θ sec² θ
Solution:
L.H.S. = tan² θ – sin² θ
= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin²θ
= sin² θ (\(\frac{1}{\cos ^{2} \theta}-1 \))
= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)
= (sin² θ) (sin² θ)sec² θ
= sin⁴ θ sec² θ
= R.H.S

xi. (sinθ + cosecθ)² + (cos θ + see θ)² = tan² θ + cot² θ + 7
Solution:
L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)²
= sin ² θ + cosec² θ + 2sinθ cosec θ
+ cos² θ + sec² θ + 2sec0 cos0
= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2
= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7
= R.H.S.

xii. sin⁸θ – cos⁸θ = (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
Solution:
L.H.S. = sin⁸θ – cos⁸θ
= (sin⁴θ)² – (cos⁴θ)²
= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)
= [(sin² θ)² – (cos² θ)² ]
. [(sin² θ)² + (cos² θ)² ]
= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] …[Y a² + b² = (a + b)² – 2ab]
= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ)
= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
= R.H.S.

xiii. sin⁶A + cos⁶A = 1 – 3 sin²A + 3sin⁴A
Soluiton:
L.H.S. = sin⁶A + cos⁶A
= (sin² A)³ + (cos² A)³
= (sin² A + cos² A)³
– 3sin²A cos²A(sin² A + cos² A)
…[ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3sin²A cos²A (1)
= 1 – 3sin²A cos²A
= 1 – 3 sin²A (1 – sin²A)
= 1 – 3 sin²A + 3sin⁴A
= R.H.S.

xiv. (1 + tanA tanB)² + (tanA – tanB)² = sec ²A sec²B
Solution:
L.H.S. = (1 + tanA tanB)² + (tanA – tanB)²
= 1 + 2tanA tanB + tan²A tan² + tan² A- 2tanA tanB + tan²B
= 1 + tan²A + tan² B + tan²A tan²B
= 1(1+ tan²A) + tan² B(1 + tan²A)
= (1 + tan²A) (1 + tan²B)
= sec²A sec²B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)
Solution:
We know that cosec²θ – cot² θ = 1
∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)
Solution:
We know that
tan²θ = sec² θ – 1
∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)
Solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)
solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)

Maharashtra State Board 11th Maths Book Solutions 

• Trigonometry – I Ex 2.1 Class 11 Maths Solutions
• Trigonometry – I Ex 2.2 Class 11 Maths Solutions
• Trigonometry – I Miscellaneous Exercise 2 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Exercise 3.2 Questions And Answers Maharashtra Board

Question 1.
Find the Cartesian co-ordinates of the point whose polar co-ordinates are :
(i) \(\left(\sqrt{2}, \frac{\pi}{4}\right)\)
Solution:
Here, r = \(\sqrt {2}\) and θ = \(\frac{\pi}{4}\)
Let the cartesian coordinates be (x, y)
Then, x = rcosθ = \(\sqrt {2}\)cos\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
y = rsinθ = \(\sqrt {2}\)sin\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
∴ the cartesian coordinates of the given point are (1, 1).

(ii) \(\left(4, \frac{\pi}{2}\right)\)
Solution:

(iii) \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\)
Solution:
Here, r = \(\frac{3}{4}\) and θ = \(\frac{3 \pi}{4}\)
Let the cartesian coordinates be (x, y)

(iv) \(\left(\frac{1}{2}, \frac{7 \pi}{3}\right)\)
Solution:
Here, r = \(\frac{1}{2}\) and θ = \(\frac{7 \pi}{4}\)
Let the cartesian coordinates be (x, y)


∴ the cartesian coordinates of the given point are \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)

Question 2.
Find the of the polar co-ordinates point whose Cartesian co-ordinates are.
(i) \((\sqrt{2}, \sqrt{2})\)
Solution:
Here x = \(\sqrt {2}\) and y = \(\sqrt {2}\)
∴ the point lies in the first quadrant.
Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (\(\sqrt {2}\) )² + (\(\sqrt {2}\) )² = 2 + 2 = 4
∴ r = 2 … [∵ r > 0]
cos θ = \(\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
and sin θ = \(\frac{y}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
∴ tan θ = 1
Since the point lies in the first quadrant and
0 ≤ θ ≤ 2π, tan θ = 1 = tan\(\frac{\pi}{4}\)
∴ θ = \(\frac{\pi}{4}\)
∴ the polar coordinates of the given point are \(\left(2, \frac{\pi}{4}\right)\).

(ii) \(\left(0, \frac{1}{2}\right)\)
Solution:
Here x = 0 and y = \(\frac{1}{2}\)
the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (0)² + \(\left(\frac{1}{2}\right)^{2}=0+\frac{1}{4}=\frac{1}{4}\)
∴ r = \(\frac{1}{2}\) …[∵ r > 0]
cosθ = \(\frac{x}{r}=\frac{0}{(1 / 2)}\) = 0
and sin θ = \(\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}\) = 1
Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π
cosθ = 0 = cos\(\frac{\pi}{2}\) and sinθ = 1 = sin\(\frac{\pi}{2}\)
∴ θ = \(\frac{\pi}{2}\)
∴ the polar coordinates of the given point are \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\).

(iii) \((1,-\sqrt{3})\)
Solution:
Here x = 1 and y = \(-\sqrt{3}\)
∴ the point lies in the fourth quadrant.
Let the polar coordinates be (r, θ).
Then, r² = x² + y² = (1)² + (\(-\sqrt {3}\) )² = 1 + 3 = 4
∴ r = 2 … [∵ r > 0]

∴ the polar coordinates of the given point are \(\left(2, \frac{5 \pi}{3}\right)\).

(iv) \(\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\)
Solution:

Question 3.
In ∆ABC, if ∠A = 45º, ∠B = 60º then find the ratio of its sides.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\)
∴ \(\frac{a}{b}=\frac{\sin A}{\sin B}\) and \(\frac{b}{c}=\frac{\sin B}{\sin C}\)
∴ a : b : c = sinA : sinB : sinC
Given ∠A = 45° and ∠B = 60°
∵ ∠A + ∠B + ∠C = 180°
∴ 45° + 60° + ∠C = 180°
∴ ∠C = 180° – 105° = 75°

Question 4.
In ∆ABC, prove that sin \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)=\left(\frac{\boldsymbol{b}-\boldsymbol{c}}{a}\right)\) cos \(\frac{A}{2}\).
Solution:
By the sine rule,

Question 5.
With usual notations prove that 2 \(\left\{a \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{A}{2}\right\}\) = a – b + c.
Solution:

Question 6.
In ∆ABC, prove that a³sin(B – C) + b³sin(C – A) + c³sin(A – B) = 0
Solution:
By the sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a³sin (B – C) + b³sin (C – A) + c³sin (A – B)
= a³(sin B cos C – cos B sin C) + b³(sinCcos A – cos C sin A) + c³(sinAcosB – cos A sin B)

= \(\frac{1}{2 k}\) [a²(a² + b² – c²) – a²(a² + c² – b²) + b²(b² + c² – a²) – b²(a² + b² – c²) + c²(c² + a² – b²) – c²(b² + c² – a²)]
= \(\frac{1}{2 k}\) [a⁴ + a²b² – a²c² – a⁴ – a²c² + a²b² + b⁴ + b²c² – a²b² – a²b² – b⁴ + b²c² + c⁴ + a²c² – b²c² – b²c² – c⁴ + a²c²]
= \(\frac{1}{2 k}\)(0) = 0 = RHS.

Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a², b², c² are also in A.P
Solution:
By the sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) =\(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb, sin C = kc …(1)
Now, cot A, cotB, cotC are in A.P.
∴ cotC – cotB = cotB – cot A
∴ cotA + cotC = 2cotB

Question 8.
In ∆ABC, if a cos A = b cos B then prove that the triangle is right angled or an isosceles traingle.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = k
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sin A cos A = k sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B ∴ sin 2A – sin 2B = 0
∴ 2 cos (A + B)∙sin (A -B) = 0
∴ 2cos (π – C)∙sin(A – B) = 0 … [∵ A + B + C = π]
∴ -2 cos C∙sin (A – B) = 0
∴ cos C = 0 OR sin(A -B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.

Question 9.
With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a² + b² + c².
Solution:
LHS = 2 (bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
= 2bc \(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\) + 2ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) + 2ab\(\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\) …(By cosine rule]
= b² + c² – a² + c² + a² – b² + a² + b² – c² = a² + b² + c² = RHS.

Question 10.
In △ABC, if a = 18, b = 24, c = 30 then find the values of
(i) cos A
Solution:
Given : a = 18, b = 24 and c = 30
∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36

(ii) sin\(\frac{A}{2}\)
Solution:

(iii) cos\(\frac{A}{2}\)
Solution:

(iv) tan\(\frac{A}{2}\)
Solution:

(v) A(△ABC)
Solution:

(iv) sin A.
Solution:

Question 11.
In △ABC prove that (b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan\(\frac{B}{2}\) = (a + b – c) tan\(\frac{C}{2}\).
Solution:
(b + c – a) tan \(\frac{A}{2}\)

Question 12.
In △ABC prove that sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\) = \(\frac{[A(\triangle A B C)]^{2}}{a b c s}\)
Solution:
LHS = sin \(\frac{A}{2}\)∙sin \(\frac{B}{2}\)∙sin \(\frac{C}{2}\)

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Trigonometric Functions Ex 3.1 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.2 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.3 Class 12 Maths Solutions
• Trigonometric Functions Miscellaneous Exercise 3 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.1 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.2 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.3 Class 12 Maths Solutions
• Pair of Straight Lines Miscellaneous Exercise 4 Class 12 Maths Solutions

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

11th Maths Part 1 Trigonometry – I Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
If 2sin A = 1 = \(\sqrt{2}\) cos B and \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos² A = 1 – sin² A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)
∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)
Since \(\frac{\pi}{2}\) < A < π
A lies in the 2nd quadrant.

We know that,
Sin² B = 1 – cos² B = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\)
∴ sin B = \(\pm \frac{1}{\sqrt{2}}\)
Since \(\frac{3 \pi}{2}\) < B < 2π
B lies in the 4th quadrant,

Question 2.
If \(\) and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)
We know that,
cos² A = 1 – sin² = 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ Cos A = ± \([{4}{5}\)
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –\(\frac{4}{5}\)
Sin B = 4/5
We know that,
cos²B = 1 – sin²B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)
∴ Cos B = ±\(\frac{4}{5}\)
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = \(\frac{1}{2}\), evaluate \(\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}\)
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = \(\frac{x}{3}\) and tan θ= \(\frac{y}{4}\)
We know that,
sec²θ – tan²θ = 1

∴ 16x² – 9y² = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ = \(\) and cot θ = \(\)
We know that,
cosec² θ – cot² θ =

16x² – 9y² = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x² + y² = (4cos θ – 5sin θ)² + (4sin θ + 5cos θ)²
= 16cos²θ – 40 sinθ cosθ + 25 sin²θ + 16 sin² θ + 40sin θ cos θ + 25 cos² θ
= 16(sin² θ + cos² θ) + 25(sin² θ + cos² θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = \(\frac{x-5}{6}\) and cot θ = \(\frac{y-3}{8}\)
We know that,
cosec² θ – cot² θ = 1
∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = \(\frac{3-2 x}{4}\) and sec θ = \(\frac{3 y-5}{3}\)θ
We know that, sec² θ – tan² θ = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

Question 5.
If 2sin² θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin² θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = \(\frac{-3}{2}\)
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
\(\sqrt{1-\cos ^{2} \theta}\) = 0 …[ ∵ sin² θ = 1- cos² θ]
∴ 1 – cos² θ = 0
∴ cos² θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos² θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos²θ – 11 cos θ + 5 = 0
∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos² θ = 3sin θ.
Solution:
2cos²0 = 3sin θ
∴ 2(1 – sin² θ) = 3sin θ
∴ 2 – 2sin² θ = 3sin θ
∴ 2sin² θ + 3sin 9-2 = θ
∴ 2sin² θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan² θ + 3 = 9sec θ
∴ 5(sec² θ – 1) + 3 = 9sec θ
∴ 5sec² θ – 5 + 3 = 9sec θ
∴ 5sec² θ – 9sec θ – 2 = 0
∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos²θ = 16(1 – sin θ)²
∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ)
∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ
∴ 25sin² θ – 32sin θ + 7 = 0
∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = \(\frac{7}{25}\)
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or \(\frac{7}{25}\)
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or \(\frac{7}{25}\).]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)
∴ \(\frac{1+\cos \theta}{\sin \theta}=5\)
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos² θ = 25 sin² θ
∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ)
∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ
∴ 26 cos² θ + 2 cos θ – 24 = 0
∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = \(\frac{24}{26}=\frac{12}{13}\)
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = \(\frac{12}{13}\)
∴ sec θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)
[Note: Answer given in the textbook is -1 or \(\frac{13}{12}\).
However, as per our calculation it is only \(\frac{13}{12}\).]

Question 11.
If cot θ = \(\frac{3}{4}\) and π < θ < \(\frac{3 \pi}{2}\), then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec²θ = 1 + cot² θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)
∴ cosec² θ = \(\frac{25}{16}\)
∴ cosec θ = \(\pm \frac{5}{4}\)
Since π < θ < \(\frac{3 \pi}{2}\)
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –\(\frac{5}{4}\)
cot θ = \(\frac{3}{4}\)
tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)
We know that,
sec² θ = 1 + tan² θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ sec θ = ±\(\frac{5}{3}\)
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –\(\frac{5}{3}\)
cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)
∴ 4cosec θ + 5cos θ
= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, \(\sqrt{3}\))
ii. (-1, -1) iv. (-\(\sqrt{3}\), 1)
Solution:
i. (x, y) = (5, 5)
∴ r = \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{25+25}\)
\(=\sqrt{50}=5 \sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = ( 1, \(\sqrt{3}\))
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)
∴ tan θ = tan \(\frac{\pi}{4}\)
Since the given point lies in the 3rd quadrant,
tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) …[∵ tan (n + x) = tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{4}\right)\)
∴ θ = \(\frac{5 \pi}{4}\) = 225°
∴ the required polar co-ordinates are (\(\sqrt{2}\), 225°).

iv. (x, y) = (-\(\sqrt{3}\) , 1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)
Since the given point lies in the 2nd quadrant,
tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) …[∵ tan (π – x) = – tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{6}\right)\)
∴ θ = \(\frac{5 \pi}{6}\) = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin\(\frac{19 \pi^{e}}{3}\)
ii. cos 1140°
iii. cot \(\frac{25 \pi^{e}}{3}\)
Solution:
i. We know that sine function is periodic with period 2π.
sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = \(\frac {1}{2}\)

iii. We know that cotangent function is periodic with period π.
cot \(\frac{25 \pi}{3}\) = cot \(\left(8 \pi+\frac{\pi}{3}\right)\) = cot \(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\)
dhana work.txt
Displaying dhana work.txt.

Maharashtra State Board 11th Maths Book Solutions 

• Trigonometry – I Ex 2.1 Class 11 Maths Solutions
• Trigonometry – I Ex 2.2 Class 11 Maths Solutions
• Trigonometry – I Miscellaneous Exercise 2 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Exercise 3.1 Questions And Answers Maharashtra Board

Question 1.
Find the principal solutions of the following equations :
(i) cos θ= \(\frac{1}{2}\)
Solution:
We know that, cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) and cos (2π – θ) = cos θ
∴ cos\(\frac{\pi}{3}\) = cos(2π – \(\frac{\pi}{3}\)) = cos\(\frac{5 \pi}{3}\)
∴ cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\) = \(\frac{1}{2}\), where
0 < \(\frac{\pi}{3}\) < 2π and 0 < \(\frac{5 \pi}{3}\) < 2π
∴ cos θ = \(\frac{1}{2}\) gives cos θ = cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\)
∴ θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)
Hence, the required principal solutions are
θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

(ii) sec θ = \(\frac{2}{\sqrt{3}}\)
Solution:

(iii) cot θ = \(\sqrt {3}\)
Solution:
The given equation is cot θ = \(\sqrt {3}\) which is same as tan θ = \(\frac{1}{\sqrt{3}}\).
We know that,

Hence, the required principal solution are
θ = \(\frac{\pi}{6}\) and θ = \(\frac{7 \pi}{6}\).

(iv) cot θ = 0.
Solution:

Question 2.
Find the principal solutions of the following equations:
(i) sinθ = \(-\frac{1}{2}\)
Solution:
We know that,
sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\) and sin (π + θ) = -sinθ,
sin(2π – θ) = -sinθ

Hence, the required principal solutions are
θ = \(\frac{7\pi}{6}\) and θ = \(\frac{11 \pi}{6}\).

(ii) tanθ = -1
Solution:
We know that,
tan\(\frac{\pi}{4}\) = 1 and tan(π – θ) = -tanθ,
tan(2π – θ) = -tanθ

Hence, the required principal solutions are
θ = \(\frac{3\pi}{4}\) and θ = \(\frac{7 \pi}{4}\).

(iii) \(\sqrt {3}\) cosecθ + 2 = 0.
Solution:

Question 3.
Find the general solutions of the following equations :
(i) sinθ = \(\frac{1}{2}\)
Solution:
(i) The general solution of sin θ = sin ∝ is
θ = nπ + (-1 )ⁿ∝, n ∈ Z
Now, sinθ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\) …[∵ sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\)]
∴ the required general solution is
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z.

(ii) cosθ = \(\frac{\sqrt{3}}{2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
Now, cosθ = \(\frac{\sqrt{3}}{2}\) = cos\(\frac{\pi}{6}\) …[∵ cos\(\frac{\pi}{6}\) = \(\frac{\sqrt{3}}{2}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z.

(iii) tanθ = \(\frac{1}{\sqrt{3}}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, tan θ = \(\frac{1}{\sqrt{3}}\) = tan\(\frac{\pi}{6}\) …[tan\(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z.

(iv) cotθ = 0.
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, cot θ = 0 ∴ tan θ does not exist
∴ tanθ = tan\(\frac{\pi}{2}\) [∵ tan\(\frac{\pi}{2}\) does not exist]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{2}\), n ∈ Z.

Question 4.
Find the general solutions of the following equations:
(i) secθ = \(\sqrt {2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = nπ ± ∝, n ∈ Z.
Now, secθ = \(\sqrt {2}\) ∴ cosθ = \(\frac{1}{\sqrt{2}}\)
∴ cosθ = cos\(\frac{\pi}{4}\) ….[cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{4}\), n ∈ Z.

(ii) cosecθ = –\(\sqrt {2}\)
Solution:
The general solution of sinθ = sin∝ is

(iii) tanθ = -1
Solution:
The general solution of tanθ = tan∝ is

Question 5.
Find the general solutions of the following equations :
(i) sin 2θ = \(\frac{1}{2}\)
Solution:
The general solution of sin θ = sin ∝ is
θ = nπ + (-1)ⁿ∝, n ∈ Z

(ii) tan \(\frac{2 \theta}{3}\) = \(\sqrt {3}\)
Solution:
The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1
Solution:
The general solution of tan θ = tan ∝ is

Question 6.
Find the general solutions of the following equations :
(i) 4 cos²θ = 3
Solution:
The general solution of cos²θ = cos² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 cos²θ = 3

(ii) 4 sin²θ = 1
Solution:
The general solution of sin²θ = sin² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 sin²θ = 3

(iii) cos 4θ = cos 2θ
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z
Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z
Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = \(\frac{n \pi}{3}\), n ∈ Z
Hence, the required general solution is
θ = \(\frac{n \pi}{3}\), n ∈ Z or ∴ θ = nπ, n ∈ Z.
Alternative Method:
cos 4θ = cos 2θ
∴ cos4θ – cos 20 = 0
∴ -2sin\(\left(\frac{4 \theta+2 \theta}{2}\right)\)∙sin\(\left(\frac{4 \theta-2 \theta}{2}\right)\) = 0
∴ sin3θ∙sinθ = 0
∴ either sin3θ = 0 or sin θ = 0
The general solution of sin θ = 0 is
θ = nπ, n ∈ Z.
∴ the required general solution is given by
3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z
i.e. θ = \(\frac{n \pi}{3}\), n ∈ Z or θ = nπ, n ∈ Z.

Question 7.
Find the general solutions of the following equations :
(i) sinθ = tanθ
Solution:
sin θ = tan θ
∴ sin θ = \(\frac{\sin \theta}{\cos \theta}\)
∴ sin θ cos θ = sin θ
∴ sin θ cos θ – sinθ = 0
∴ sin θ (cos θ – 1) = θ
∴ either sinθ = 0 or cosθ – 1 = 0
∴ either sin θ = 0 or cos θ = 1
∴ either sinθ = 0 or cosθ = cosθ …[∵ cos0 = 1]
The general solution of sinθ = 0 is θ = nπ, n ∈ Z and cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = 2nπ ± 0, n ∈ Z
∴ θ = nπ, n ∈ Z or θ = 2nπ, n ∈ Z.

(ii) tan³θ = 3tanθ
Solution:
tan³θ = 3tanθ
∴ tan³θ – 3tanθ = 0
∴ tan θ (tan²θ – 3) = 0
∴ either tan θ = 0 or tan²θ – 3 = 0
∴ either tanθ = 0 or tan²θ = 3
∴ either tan θ = 0 or tan²θ = (\(\sqrt {3}\) )³
∴ either tan θ = 0 or tan²θ = (tan\(\frac{\pi}{3}\))³ …[tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ either tanθ = 0 or tan²θ = tan²\(\frac{\pi}{3}\)
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan²θ = tan²∝ is θ = nπ ± ∝, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) cosθ + sinθ = 1.
Solution:
cosθ + sinθ = 1

Question 8.
Which of the following equations have solutions ?
(i) cos 2θ = -1
Solution:
cos 2θ = -1
Since -1 ≤ cos θ ≤ 1 for any θ,
cos 2θ = -1 has solution.

(ii) cos²θ = -1
Solution:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5
Solution:
3tanθ = 5 ∴ tanθ = \(\frac{5}{3}\)
This is possible because tan θ is any real number.
∴ 3tanθ = 5 has solution.

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Trigonometric Functions Ex 3.1 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.2 Class 12 Maths Solutions
• Trigonometric Functions Ex 3.3 Class 12 Maths Solutions
• Trigonometric Functions Miscellaneous Exercise 3 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.1 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.2 Class 12 Maths Solutions
• Pair of Straight Lines Ex 4.3 Class 12 Maths Solutions
• Pair of Straight Lines Miscellaneous Exercise 4 Class 12 Maths Solutions

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.1 Questions and Answers.

11th Maths Part 1 Trigonometry – I Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, – 30°, – 45°, – 60°, – 90°, – 120°, – 225°, – 240°, – 270°, – 315°
Solution:
Angle of measure 0°:

Let m∠XOA = 0° = 0^c
Its terminal arm (ray OA) intersects the standard
unit circle in P(1, 0).
Hence,x = 1 and y = 0
sin 0° = y = 0,
cos 0° = x = 1,
tan 0° = \(\frac{y}{x}=\frac{0}{1}\) = 0
cot 0° = \(\frac{x}{y}=\frac{1}{0}\) which is not defined
sec 0° = \(\frac{1}{x}=\frac{1}{1}\) = 1
cot 0° = \(\frac{1}{y}=\frac{1}{0}\) which is not defined,

Angle of measure 30°:
Let m∠XOA = 30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y)
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{1}{\sqrt{2}}\) and
y = PM = \(\frac{1}{\sqrt{2}}\)
∴ P = (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

Angle of measure 60°:
Let m∠XOA = 60°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Angle of measure 150°:
Let m∠XOA = 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 180°:
Let m∠XOA = 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(-1, 0).
∴ x = – 1 and y = 0
sin 180° =y = 0
cos 180° = x = -1

tan 180° = \(\frac{y}{x}\)
= \(\frac{0}{-1}\) = 0
Cosec 180° = \(\frac{1}{y}\)
= \(\frac{1}{0}\)
which is not defined.
sec 180°= \(\frac{1}{x}=\frac{1}{-1}\) = -1
cot 180° = \(\frac{x}{y}=\frac{-1}{0}\) , which is not defined.

Angle of measure 210°:
Let m∠XOA = 210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 3rd quadrant, x < 0,y < 0
∴ x = -OM = \(\frac{-\sqrt{3}}{2}\) and y = -PM = \(\frac{-1}{2}\)
∴ P ≡( \(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\) )

Angle of measure 300°:
Let m∠XOA = 300°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0,y > 0
x = OM = \(\frac{1}{2}\) = and y = -PM = \(\frac{-\sqrt{3}}{2}\)
sin 300° = y = \(\frac{-\sqrt{3}}{2}\)
cos 300° = x = \(\frac{1}{2}\)
tan 300° = \(\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\)

Angle of measure 330°:
Let m∠XOA = 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 4th quadrant, x > 0, y < 0

Angle of measure 30°
Let m∠XOA = -30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60 — 90° triangle.
op = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our calculation it is \(-\frac{1}{\sqrt{2}}\) ]

Angle of measure (-60°):
Let m∠XOA = -60°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 4’ quadrant,
x > 0, y < 0
x = OM =\(\frac{1}{2}\) and y = -PM = \(-\frac{\sqrt{3}}{2}\)

Angle of measure (-90°):
Let m∠XOA = -90°
It terminal arm (ray OA) intersects the standard unit circle at P(0, -1)
∴ x = 0 and y = -1
sin (-90°) = y = -1
cos (-90°) = s = 0

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (- 225°):
Let m∠XOA = – 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 2400):
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30°  – 60° –  900 triangle.

Since point P lies in the 2nd quadrant, x<0, y>0

Angle of measure (- 270°):
Let m∠XOA = – 270°
Its terminal arm (ray OA)
intersects the standard unit,
circle at P(0, 1).
∴ x = 0 and y = 1
sin (- 270°) = y = 1
cos (- 270°) = x = 0
tan(-270°)= \(\frac{y}{x}=\frac{1}{0}\)
which is not defined.

Angle of measure ( 315°):
Let m∠XOA 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Question 2.
State the signs of:
i. tan 380°
ii. cot 230°
iii 468°
Solution:
1. 380° = 360° + 20°
∴ 380° and 20° are co-terminal angles.
Since 0° < 20° <90°0,
20° lies in the l quadrant.
∴ 380° lies in the 1st quadrant,
∴ tan 380° is positive.

ii. Since, 180° <230° <270°
∴ 230° lies in the 3rd quadrant.
∴ cot 230° is positive.

iii. 468° = 360°+108°
∴ 468° and 108° are co-terminal angles.
Since 90° < 108° < 180°,
108° lies in the 2nd quadrant.
∴ 468° lies in the 2nd quadrant.
∴ sec 468° is negative.

Question 3.
State the signs of cos 4^c and cos 4°. Which of these two functions is greater?
Solution:
Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)
Since 1^c = 57° nearly,
180° < 4^c < 270°
∴ 4^c lies in the third quadrant.
∴ cos 4^c < 0 ………(ii)
From (i) and (ii),
cos 4° is greater.

Question 4.
State the quadrant in which 6 lies if
i. sin θ < 0 and tan θ > 0
ii. cos θ < 0 and tan θ > 0
Solution:
i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

Question 5.
Evaluate each of the following:
i. sin 30° + cos 45° + tan 180°
ii. cosec 45° + cot 45° + tan 0°
iii. sin 30° x cos 45° x lies tan 360°
Solution:
i. We know that,
sin30° = 1/2, cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 180° = 0
sin30° + cos 45° +tan 180°
= \(\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2}\)

ii. We know that,
cosec 45° = \(\sqrt{2}\) , cot 45° = 1, tan 0° = 0
cosec 45° + cot 45° + tan 0°
= \(\sqrt{2}\) + 1 + 0 = \(\sqrt{2}\) + 1

iii. We know that,
sin 30° = \(\frac{1}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 360° = 0
sin 30° x cos 45° x tan 360°
= \(\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\) = 0

Question 6.
Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, – 4).
Solution:
Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).
∴ x = 3 and y = -4
r = OP

Question 7.
If cos θ = \(\frac{12}{13}, 0<\theta<\frac{\pi}{2}\) find the value of \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^{2} \theta}\)
Solution:
cos θ = \(\frac{12}{13}\)
We know that,
sin² θ = 1 – cos²θ

∴ sin θ = ± \(\frac{5}{13}\)
Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, ∴ sin θ > 0

Question 8.
Using tables evaluate the following:
i. 4 cot 45° – sec2 60° + sin 30°
ii.\(\cos ^{2} 0+\cos ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{2}\)
Solution:
i. We know that,
cot 45° = 1, sec 60° = 2, sin 30° = 1/2
4 cot 45° – sec² 60° + sin 30°
= 4(1) – (2)² + \(\frac{1}{2}\)
= 4 – 4 + \(\frac{1}{2}=\frac{1}{2}\)

ii. We know that,

Question 9.
Find the other trigonometric functions if
i. cot θ = \(-\frac{3}{5}\), and 180 < θ < 270
ii. Sec A = \(-\frac{25}{7}\) and A lies in the second quadrant.
iii cot x = \(\frac{3}{4}\), x lies in the third quadrant.
iv. tan x = \(\frac{-5}{12}\) x lies in the fourth quadrant.
Solution:
i. cot θ = \(-\frac{3}{5}\)
we know that,
sin²θ = 1 – cos²θ
= 1 – \(\left(-\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ sin θ = ± \(\frac{4}{5}\)
Since 180° < 0 < 270°,
θ lies in the 3rd quadrant.
∴ sin θ < 0

Since A lies in the 2nd quadrant,
tan A < 0

iii. Given, cot x = \(\frac{3}{4}\)
We know that,
cosec² x = 1 + cot² x
= 1 + \(\left(\frac{3}{4}\right)^{2}=1+\frac{9}{16}=\frac{25}{16}\)
∴ cosec x = ± \(\frac{5}{4}\)
Since x lies in the 3rd quadrant, cosec x < 0

iv. Given, tan x = \(-\frac{5}{12}\)
sec² x = 1 + tan²
= 1 + \(\left(-\frac{5}{12}\right)^{2}\)
= 1 + \(\frac{25}{144}=\frac{169}{144}\)
∴ sec x = ± \(\frac{13}{12}\)
Since x lies in the 4th quadrant,
sec x > 0

Maharashtra State Board 11th Maths Book Solutions 

• Trigonometry – I Ex 2.1 Class 11 Maths Solutions
• Trigonometry – I Ex 2.2 Class 11 Maths Solutions
• Trigonometry – I Miscellaneous Exercise 2 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2B Questions and Answers.

12th Maths Part 1 Matrices Miscellaneous Exercise 2B Questions And Answers Maharashtra Board

I) Choose the correct answer from the given alternatives in each of the following questions :
Question 1.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\), adj = \(\left(\begin{array}{ll}
4 & a \\
-3 & b
\end{array}\right)\) then the values of a and b are,
(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

Question 2.
The inverse of \(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\) is

Solution:
\(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\)

Question 3.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right)\) and A(adj A) = k 1, then the value of k is
(a) 1
(b) -1
(c) 0
(d) -3
Solution:
(d) -3 [Hint : A(adj A) = |A| ∙ I]

Question 4.
If A = \(\left(\begin{array}{ll}
2 & -4 \\
3 & 1
\end{array}\right)\), then the adjoint of matrix A is

Solution:\(\left(\begin{array}{ll}
1 & 4 \\
-3 & 2
\end{array}\right)\)

Question 5.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\) and A(adj A) = kI, then the value of k is
(a) 2
(b) -2
(c) 10
(d) -10
Solution:
(b) -2

Question 6.
If A = \(\left(\begin{array}{rr}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right)\), then A^-1 does not exist if λ = ………..
(a) 0
(b) ± 1
(c) 2
(d) 3
Solution:
(b) ± 1

Question 7.
If A = \(\left[\begin{array}{ll}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) then A^-1 = ….

Solution:
\(\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)

Question 8.
If F (∝) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\) where ∝ ∈ R then [F(∝)]^-1 is =
(a) F(-∝)
(b) F(∝^-1)
(c) F(2∝)
(d) None of these
Solution:
(a) F(-∝)

Question 9.
The inverse of A = \(\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\)
(a) I
(b) A
(c) A’
(d) -I
Solution:
(b) A

Question 10.
The inverse of a symmetric matrix is
(a) Symmetric
(b) Non-symmetric
(c) Null matrix
(d) Diagonal matrix
Solution:
(a) Symmetric

Question 11.
For a 2 × 2 matrix A, if A(adjA) = \(\left(\begin{array}{ll}
10 & 0 \\
0 & 10
\end{array}\right)\) then determinant A equals
(a) 20
(b) 10
(c) 30
(d) 40
Solution:
(b) 10

Question 12.
If A² = \(-\frac{1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]\) then A =

Solution:
\(-\frac{1}{2}\left[\begin{array}{cc}
2 & 4 \\
1 & 1
\end{array}\right]\)

II) Solve the following equations by the methods of inversion.
(i) 2x – y = -2 , 3x + 4y = 5
Solution:
The given equations can be written in the matrix form as :



By equality of matrices,
x = \(-\frac{5}{11}\), y = \(\frac{12}{11}\) is the required solution.

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:
The given equations can be written in the matrix form as :

= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:
The given equations can be written in the matrix form as :

= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A^-1 exists.
Now, we have to find the cofactor matrix


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B
∴ X = \(\frac{1}{51}\left[\begin{array}{rrr}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]\)

By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:
The given equations can be written in the matrix form as :

= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B

∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

Question 2.
Express the following equation in matrix from and solve them by the method of reduction.
(i) x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x – y + z = 1 ……(1)
y – 2z = -1 …..(2)
5z = 5 ….(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y – 2 = -1 ∴ y = 1
Substituting y = 1, z = 1 in (1), we get,
x – 1 + 1 = 1
∴ x = 1
Hence, x = 1, y = 1, z = 1 is the required solution.

(ii) x + y = 1, y + z = \(\frac{5}{3}\), z + x = \(\frac{4}{3}\).
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + y = 1 ……(1)
y + z = \(\frac{5}{3}\) …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = \(\frac{5}{3}\) ∴ y = \(\frac{2}{3}\)
Substituting y = \(\frac{2}{3}\) in (1), we get,
x + \(\frac{2}{3}\) = 1 ∴ x = \(\frac{1}{3}\)
Hence, x = \(\frac{1}{3}\), y = \(\frac{2}{3}\), z = 1 is the required solution.

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as :

∴ \(\left[\begin{array}{r}
x+2 y+3 z \\
0-5 y-5 z \\
0+0-8 z
\end{array}\right]\) = \(\left[\begin{array}{r}
8 \\
-15 \\
-8
\end{array}\right]\)
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) x + 2y + z = 8, 2x + 3y – z =11 and 3x – y – 2z = 5
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + 2y + z = 8 … (1)
-y – 3z = -5 … (2)
16z = 16 … (3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-y – 3 = -5, ∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 1 = 8 ∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x + 3y + 2z = 6 …(1)
y + \(\frac{3}{2}\) z = 4 …(2)
\(\frac{31}{2}\)z = 31 …..(3)
From (3), z = 2
Substituting z = 2 in (2), we get,
y + \(\frac{3}{2}\)z = 4
y + \(\frac{3}{2}\)(2) = 4
y + 3 = 4
y = 1
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.

Question 3.
The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 6.
3z + y = 11, i.e., y + 3z = 11 and x + z = 2y,
i.e., x – 2y + z = 0
Hence, the system of the linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in the matrix form as :

By equality of matrices,
x + y + z = 6 …(1)
y + 3z = 11 …(2)
-3y = -6 …(3)
From (3), y = 2
Substituting y = 2 in (2), we get,
2 + 3z = 11
∴ 3z = 9 ∴ z = 3
Put y = 2, z — 3 in (1), we get,
x + 2 + 3 = 6 ∴ x = 1
∴ x = 1, y = 2, z = 3
Hence, the required numbers are 1, 2 and 3.

Question 4.
The cost of 4 pencils, 3 pens and 2 books is ₹ 150. The cost of 1 pencil, 2 pens and 3 books is ₹ 125. The cos of 6 pencils, 2 pens and 3 books is ₹ 175. Fild the cost of each item by using Matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 book be ₹x, ₹ y, ₹ z respectively.
According to the given conditions,
4x + 3y + 2z = 150
x + 2y + 3z = 125
6x + 2y + 3z = 175
The equations can be written in matrix form as :

By equality of matrices,
x + 2y + 3z = 125 …(1)
-5y – 10z = -350 …(2)
5z = 125 …(3)
From (3), z = 25
Substituting z = 25 in (2), we get
-5y – 10(25) = -350
∴ -5y = -350 + 250 = -100
∴ y = 20
Substituting y = 20, z = 25 in (1), we get
x + 2(20) + 3(25) = 125
∴ x = 125 – 40 – 75 = 10
∴ x = 10, y = 20, z = 25
Hence, the cost of 1 pencil is ₹ 10, 1 pen is ₹ 20 and 1 book is ₹ 25.

Question 5.
The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times first number to the sum of second and third number, we get 12. Find the three numbers by using Matrices.
Solution:
Let the numbers be x, y and z.
According to the given conditions,
x + y + z = 6
3z + x = 7, i.e., x + 3z = 7
and 3x + y + z = 12
Hence, the system of linear equations is
x + y + z = 6
x + 3z = 7
3x + y + z = 12
These equations can be written in matrix form as :

By equality of matrices,
x + y + z = 6 …(1)
-y + 2z = 1 …(2)
-3y = -5 …(3)
From (3), y = \(\frac{5}{3}\)
Substituting y = \(\frac{5}{3}\) in (2), we get,
–\(\frac{5}{3}\) + 2z = 1
∴ 2z = 1 + \(\frac{5}{3}\) = \(\frac{8}{3}\)
∴ z = \(\frac{4}{3}\)
Substituting y =\(\frac{5}{3}\), z = \(\frac{5}{3}\) in (1), we get,
x + \(\frac{5}{3}+\frac{4}{3}\) = 6
∴ x = 3
∴ x = 3, y = \(\frac{5}{3}\), z = \(\frac{4}{3}\)
Hence, the required numbers are 3, \(\frac{5}{3}\) and \(\frac{4}{3}\).

Question 6.
The sum of three numbers is 2. If twice the second number is added to the sum of first and third number, we get 1 adding five times the first number to the sum of second and third we get 6. Find the three numbers by using matrices.
Solution:
Let the three numbers be x, y and z.
According to the question,
x + y + 2
x + 2y + z = 1
5x + y + z = 6
The given system of equations can be written in matrix form as follows:

Question 7.
An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from these investimest is ₹ 350. If the total annual income from first two investment is ₹ 70 more than the income from the third, find the amount of each investment using matrix method.
Solution:
Let the amounts in three investments by ₹ x, ₹ y and ₹ z respectively.
Then x + y + z = 5000
Since the rate of interest in these investments are 6%, 7% and 8% respectively, the annual income of the three investments are \(\frac{6 x}{100}\), \(\frac{7 y}{100}\) and \(\frac{8 z}{100}\) respectively.
According to the given conditions,
\(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\) = 350
i.e. 6x + 7y + 8z = 35000
Also, \(\frac{6 x}{100}+\frac{7 y}{100}\) = \(\frac{8 z}{100}\) + 70
i.e. 6x + 7y – 8z = 7000
Hence, the system of linear equation is
x + y + z = 5000
6x + 7y + 8z = 35000
6x + 7y – 8z = 7000
These equations can be written in matrix form as :


By equality of matrices,
x + y + z = 5000 …(1)
y + 2z = 5000 …(2)
-16z = -28000 ….(3)
From (3), z = 1750
Substituting z = 1750 in (2), we get,
y + 2(1750) = 5000
∴ y = 5000 – 3500 = 1500
Substituting y = 1500, z = 1750 in (1), we get,
x + 1500 + 1750 = 5000
∴ x = 5000 – 3250 = 1750
∴ x = 1750, y = 1500, z = 1750
Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.

Question 8.
The sum of the costs of one ook each of Mathematics, Physics and Chemistry is ₹ 210. Total cost of a mathematics book, 2 physics books, and a chemistry book is ₹ 240 Also the total cost of a Mathematics book, 3 physics book and chemistry books is Rs. 300/-. Find the cost of each book, using Matrices.
Solution:

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Mathematical Logic Ex 1.1 Class 12 Maths Solutions
• Mathematical Logic Ex 1.2 Class 12 Maths Solutions
• Mathematical Logic Ex 1.3 Class 12 Maths Solutions
• Mathematical Logic Ex 1.4 Class 12 Maths Solutions
• Mathematical Logic Ex 1.5 Class 12 Maths Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Maths Solutions
• Matrices Ex 2.1 Class 12 Maths Solutions
• Matrices Ex 2.2 Class 12 Maths Solutions
• Matrices Ex 2.3 Class 12 Maths Solutions
• Matrices Miscellaneous Exercise 2A Class 12 Maths Solutions
• Matrices Miscellaneous Exercise 2 Class 12 Maths Solutions

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Miscellaneous Exercise 1 Questions and Answers.

11th Maths Part 1 Angle and its Measurement Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

I. Select the correct option from the given alternatives.

Question 1.
\(\left(\frac{22 \pi}{15}\right)^{c}x\) is equal to
(A) 246°
(B) 264°
(C) 224°
(D) 426°
Answer:
(B) 264°

Question 2.
156° is equal to

Answer:
(B)

Question 3.
A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 metres when it traces the angle of 12° at the centre, then the length of the rope is
(A) 70 m
(B) 55 m
(C) 40 m
(D) 35 m
Answer:
(A) 70 m

Question 4.
A pendulum 14 cm long oscillates through an angle of 12°, then the angle of the path described by its extremities is


Answer:
(D)

Question 5.
Angle between hands of a clock when it shows the time 9 :45 is
(A) (7.5)°
(B) (12.5)°
(C) (17.5)°
(D) (22.5)°
Answer:
(D) (22.5)°

Question 6.
20 metres of wire is available for fencing off a flower-bed in the form of a circular sector of radius 5 metres, then .the maximum area (in sq. m.) of the flower-bed is
(A) 15
(B) 20
(C) 25
(D) 30
Answer:
(C) 25
r + r + rθ = 20m
2r + rθ = 20

Question 7.
If the angles of a triangle are in the ratio 1:2:3, then the smallest angle in radian is
(A) \(\frac{\pi}{3}\)
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{2}\)
(D) \(\frac{\pi}{9}\)
Answer:
(B) \(\frac{\pi}{6}\)

Question 8.
A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find the ratio of their areas?
(A) 5:1
(B) 4:5
(C) 5:4
(D) 3:4
Answer:
(B) 4:5

Question 9.
Find the measure of the angle between hour- hand and the minute hand of a clock at twenty minutes past two.
(A) 50°
(B) 60°
(C) 54°
(D) 65°
Answer:
(A) 50°

Question 10.
The central angle of a sector of circle of area 9π sq.cm is 60°, the perimeter of the sector is
(A) π
(B) 3 + π
(C) 6 + π
(D) 6
Answer:
(C) 6 + π

II. Answer the following.

Question 1.
Find the number of sides of a regular polygon, if each of its interior angles is \(\frac{3 \pi^{c}}{4}\).
Solution:
Each interior angle of a regular polygon
= \(\frac{3 \pi}{4}=\left(\frac{3 \pi}{4} \times \frac{180}{\pi}\right)^{\circ}\) = 135°
Interior angle + Exterior angle = 180°
∴ Exterior angle = 180° – 135° = 45°
Let the number of sides of the regular polygon be n.
But in a regular polygon, exterior angle = \(\frac{360^{\circ}}{\text { no.of sides }}\)
∴ 45° = \(\frac{360^{\circ}}{\mathrm{n}}\)
∴ n = \(\frac{360^{\circ}}{45^{\circ}}\) = 8
∴ Number of sides of a regular polygon = 8.

Question 2.
Two circles each of radius 7 cm, intersect each other. The distance between their centres is 7√2 cm. Find the area common to both the circles.
Solution:
Let O and O₁ be the centres of two circles intersecting each other at A and B.
Then OA = OB = O₁A = O₁B = 7 cm
and OO₁ = 7√2 cm
OO₁² = 98 ………………(i)
Since OA² + O₁A² = 7²
= 98
= OO₁² …..[ from (i)]
m∠OAO₁ = 90°
□ OAO₁B is a square.
m∠AOB = m∠AO₁B = 90°

A(□ OAO₁B) = (side)² = (7)² = 49 sq.cm
∴ Required area = area of shaded portion = A(sector OAB) + A(sector O₁AB)) – A(□ OAO₁B)

Question 3.
∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.
Solution:
Let ‘O’ be the centre of the circle drawn on QR as a diameter.
Let the circle intersect seg PQ and seg PR at points M and N respectively.
Since l(OQ) = l(OM),
m∠OM Q = m∠OQM = 60°
m∠MOQ = 60°
Similarly, m∠NOR = 60°
Given, QR =18 cm.
r = 9 cm

θ = 60° = (60 x \(\frac{\pi}{180}\))^c
= \(\left(\frac{\pi}{3}\right)^{c}\)
∴ l(arc MN) = S = rθ = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 4.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm.
Solution:
Let S be the length of the arc and r be the radius of the circle.
θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
S = 37.4 cm
Since S = rθ,

Question 5.
A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?
Solution:
S = 10 cm and r = 4 cm
Since S = rθ,
10 = 4 x θ

Question 6.
If two arcs of the same length in two circles subtend angles 65° and 110° at the centre. Find the ratio of their radii.
Solution:
Let r₁ and r₂ be the radii of the two circles and let their arcs of same length S subtend angles of 65° and 110° at their centres.
Angle subtended at the centre of the first circle,

Angle subtended at the centre of the second circle,

Question 7.
The area of a circle is 81TH sq.cm. Find the length of the arc subtending an angle of 300° at the centre and also the area of corresponding sector.
Solution:
Area of circle = πr²
But area is given to be 81 n sq.cm
∴ πr² = 81π
∴ r2 = 81
∴ r = 9 cm
θ = 300° = \(=\left(300 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{5 \pi}{3}\right)^{\mathrm{c}}\)
Since S = rθ
S = 9 x \(\frac{5 \pi}{3}\) = 15π cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 9 x 15π = \(\frac{135 \pi}{2}\) sq.cm

Question 8.
Show that minute-hand of a clock gains 5° 30′ on the hour-hand in one minute.
Solution:
Angle made by hour-hand in one minute
\(=\frac{360^{\circ}}{12 \times 60}=\left(\frac{1}{2}\right)^{\circ}\)
Angle made by minute-hand in one minute = \(\frac{360^{\circ}}{60}\) = 6°
∴ Gain by minute-hand on the hour-hand in one minute
= \(6^{\circ}-\left(\frac{1}{2}\right)^{\circ}=\left(5 \frac{1}{2}\right)^{\circ}\) = 5°30′
[Note: The question has been modified.]

Question 9.
A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.
Solution:
r = 1km = 1000m
l(Arc covered by train in 30 seconds)
= 30 x \(\frac{36000}{60 \times 60}\)m
∴ S = 300 m
Since S = rθ,
300 = 1000 x θ

= (17.18)°
= 17° +(0.18)°
= 17° + (0.18 x 60)’ = 17° + (10.8)’
∴ θ = 17°11′(approx.)

Question 10.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:

Let ‘O’ be the centre of the circle and AB be the chord of the circle.
Here, d = 40 cm
∴ r = \(\frac{40}{2}\) = 20 cm
Since OA = OB = AB,
∆OAB is an equilateral triangle.
The angle subtended at the centre by the minor
arc AOB is θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
= l(minor arc of chord AB) = rθ = 20 x \(\frac{\pi}{3}\)
= \(\frac{20 \pi}{3}\) cm

Question 11.
The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radians.
Solution:
Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0
∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360°
… [Sum of the angles of a quadrilateral is 360°]
∴ 4a = 360°
∴ a = 90°
According to the given condition, the greatest angle is double the least,
∴ a + 3d = 2.(a – 3d)
∴ 90° + 3d = 2.(90° – 3d)
∴ 90° + 3d = 180° – 6d 9d = 90°
∴ d = 10°
∴ The measures of the angles in degrees are
a – 3d = 90° – 3(10°) = 90° – 30° = 60°,
a – d = 90° – 10° = 80°,
a + d = 90°+ 10°= 100°,
a + 3d = 90° + 3(10°) = 90° + 30° = 120°

Maharashtra State Board Class 11 Maths Solutions 

• Angle and its Measurement Ex 1.1 Class 11 Maths Solutions
• Angle and its Measurement Ex 1.2 Class 11 Maths Solutions
• Angle and its Measurement Miscellaneous Exercise 1 Class 11 Maths Solutions

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2A Questions and Answers.

12th Maths Part 1 Matrices Miscellaneous Exercise 2A Questions And Answers Maharashtra Board

Question 1.
If A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\) then reduce it to I₃ by using column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right|\)
= 1(1 – 0) – 0 + 0 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By C₁ – 2C₂, we get, A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
-3 & 3 & 1
\end{array}\right]\)
By C₁ + 3C₃ and C₂ – 3C₃, we get,
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 2.
If A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\), then reduce it to I₃ by using row transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 2 (0 – 1) – 1(1 – 1) + 3 (1 – 0)
= -2 – 0 + 3 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\)
By R₁ – R₂, we get,

By R₁ – R₃ and By R₂ – R₃, we get
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 3.
Check whether the following matrices are invertible or not:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exists.

(ii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right|\) = 1 – 1 = 0.
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right|\) = 3 – 6 = -3 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(iv) \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right|\) = 30 – 30 = 0.
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(v) \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{cc}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\)
= sec²θ – tan²θ = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(vii) \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Solution:
let A = \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right|\)
= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)
= 15 – 20 + 9 = 4 ≠ 0
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(viii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right|\)
= 1 (-3 -6) – 2 (6 – 3) + 3 (4 + 1)
= -9 – 6 + 15 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(ix) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right|\)
= 1(32 – 30) – 2(24 – 20) + 3(18 – 16)
= 2 – 8 + 6 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

Question 4.
Find AB, if A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & -2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -1 \\
1 & 2 \\
1 & -2
\end{array}\right]\) Examine whether AB has inverse or not.
Solution:

∴ A is a non-singular matrix.
Hence, (AB)^-1 exist.

Question 5.
If A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is a nonsingular matrix then find A^-1 by elementary row transformations.
Hence, find the inverse of \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\)
Solution:
Since A is a non-singular matrix, then find A^-1 by using elementary row transformations.
We write AA^-1 = I

Comparing \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) with \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\),
we get, x = 2, y = 1, z = -1
∴ \(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{1}\) = 1, \(\frac{1}{z}\) = \(\frac{1}{-1}\) = -1
\(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) is \(\left(\begin{array}{rrr}
\frac{1}{2} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right)\).

Question 6.
if A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and X is a 2 × 2 matrix such that AX = I , then find X.
Solution:
We will reduce the matrix A to the identity matrix by using row transformations. During this pro¬cess, I will be converted to the matrix X.
We have AX = I.

Question 7.
Find the inverse of each of the following matrices (if they exist).
(i) \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right|\) = 3 + 2 = 5 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ii) \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right|\) = -2 – 1 = -3 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iii) \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right|\) = 7 – 6 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iv) \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right|\) = 14 + 15 = 29 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(v) \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right|\) = 8 – 7 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vi) \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right|\) = -21 + 20 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vii) \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right|\)
= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)
= 20 – 15 – 30 = -25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(viii) \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right|\)
= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)
= 25 + 30 -30
= 25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ix) \(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A =\(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right|\)
= 2(3 – 0) – 0 – 1(5 – 0)
= 6 – 0 – 5 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

(x) \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
= 1\(\left|\begin{array}{ll}
-2 & 1 \\
3 & 0
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & 1 \\
-1 & 1
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & -2 \\
-1 & 3
\end{array}\right|\)
|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)
= -3 – 2 + 4
= -1 ≠ 0
∴ A^-1 exists.
We have
AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & 6 & 2 \\
1 & 2 & 1 \\
2 & 5 & 2
\end{array}\right]\)

Question 8.
Find the inverse of A = \(\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right]\) by
(i) elementary row transformations
Solution:
|A| = \(\left|\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right|\)
= cosθ (cosθ – 0) + sinθ (sinθ – 0) + 0
= cos²θ + sin²θ = 1 ≠ 0
∴ A^-1 exists.
(i) Consider AA^-1 = I

(ii) elementary column transformations
Solution:
Consider A^-1A = I

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]\) find AB and (AB)^-1. Verify that (AB)^-1 = B^-1A^-1
Solution:
AB = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\) \(\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]\)




From (1) and (2), (AB)^-1 = B^-1 ∙ A^-1.

Question 10.
If A = \(\left[\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right]\), then show that A^-1 = \(\frac{1}{6}\)(A – 5I)
Solution:
|A| = \(\left|\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right|\) = 4 – 10 = -6 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 11.
Find matrix X such that AX = B, where A = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
0 & 1 \\
2 & 4
\end{array}\right]\)
Solution:
AX = B

Question 12.
Find X, if AX = B where A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\).
Solution:
AX = B

Question 13.
If A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
4 & 1 \\
3 & 1
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\) then find matrix X such that AXB = C.
Solution:
AXB = C
∴ \(\left(\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right)(\mathrm{XB})\) =\(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\)
First we perform the row transformations.

Question 14.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by adjoint method.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]_3×3 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = \(\left|\begin{array}{ll}
1 & 5 \\
4 & 7
\end{array}\right|\) = 7 – 20 = -13
A₁₂ = (-1)¹⁺²M₁₂ = \(\left|\begin{array}{ll}
1 & 5 \\
2 & 7
\end{array}\right|\) = -(7 – 10) = 3

Question 15.
Find the inverse of \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by adjoint method.
Solution:
where A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\)
|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)
|A| = -4 – 2
|A| = -6 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]3×3, where A_ij = (-1)^i+jM_ij

Question 16.
Find A^-1 by adjoint method and by elementary transformations if A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right|\)
= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)
= 0 + 12 – 9 = 3 ≠ 0
∴ A^-1 exists.
A^-1by adjoint method :
We have to find the cofactor matrix
= [A_ij]_3×3, where A_ij = (-1)^i+j M_ij

Question 17.
Find the inverse of A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by elementary column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right|\)
= 1 (2 – 6) – 0 + 1 (0 – 2)
= -4 – 2= -6 ≠ 0
∴ A^-1 exists.
Consider A^-1A = I
∴ A^-1\(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
By C₃ – C₁, we get,

Question 18.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by elementary row transformations.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 19.
Show with usual notations that for any matrix A = [a_ij]_3×3
(i) a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
(i) A₂₁ = (-1)²⁺¹M₂₁ = \(-\left|\begin{array}{ll}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|\)
= -(a₁₂a₃₃ – a₁₃a₃₂)
= -a₁₂a₃₃ + a₁₃a₃₂
A₂₂ = (-1)²⁺²M₂₂ = \(\left|\begin{array}{ll}
a_{11} & a_{13} \\
a_{31} & a_{33}
\end{array}\right|\)
= a₁₁a₃₃ – a₁₃a₃₁
A₂₃ = (-1)²⁺³M₂₃ = \(-\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{31} & a_{32}
\end{array}\right|\)
= -(a₁₁a₃₂ – a₁₂a₃₁)
= -a₁₁a₃₂+ a₁₂a₃₁
∴ a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃
= a₁₁(-a₁₂₃₃ + a₁₃a₃₂) + a₁₂(a₁₁a₃₃ – a₁₃a₃₁) + a₁₃(-a₁₁a₃₂ + a₁₂a₃₁)
= -a₁₁a₁₂a₃₃ + a₁₁a₁₃a₃₂ + a₁₁a₁₂a₃₃ – a₁₂a₁₃a₃₁ – a₁₁a₁₃a₃₂ + a₁₂a₁₃a₃₁
= 0

(ii) a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = |A|
Solution:

Question 20.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\), then find a matrix X such that XA= B.
Solution:
Consider XA = B

12th Maharashtra State Board Maths Solutions Pdf Part 1

• Mathematical Logic Ex 1.1 Class 12 Maths Solutions
• Mathematical Logic Ex 1.2 Class 12 Maths Solutions
• Mathematical Logic Ex 1.3 Class 12 Maths Solutions
• Mathematical Logic Ex 1.4 Class 12 Maths Solutions
• Mathematical Logic Ex 1.5 Class 12 Maths Solutions
• Mathematical Logic Miscellaneous Exercise 1 Class 12 Maths Solutions
• Matrices Ex 2.1 Class 12 Maths Solutions
• Matrices Ex 2.2 Class 12 Maths Solutions
• Matrices Ex 2.3 Class 12 Maths Solutions
• Matrices Miscellaneous Exercise 2A Class 12 Maths Solutions
• Matrices Miscellaneous Exercise 2 Class 12 Maths Solutions