Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.5

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.5

Question 1.
In Δ ABC, A + B + C = π, show that
cos2A + cos2B + cos2C = – 1 – 4 cosA cosB cosC
Solution:
L.H.S. = cos 2A + cos 2B + cos 2C
= \(2 \cdot \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cdot \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)+\cos 2 \mathrm{C}\)
= 2.cos(A + B).cos (A – B) + 2cos²C – 1
In ΔABC, A + B + C = π
∴ A + B = π – C
∴ cos(A + B) = cos(π – C)
∴ cos(A + B) = – cosC ………….(i)

∴ L.H.S. = – 2.cos C.cos (A – B) + 2.cos²C – 1 …[From(i)]
= – 1 – 2.cosC.[cos(A – B) – cosC]
= – 1 – 2.cos C.[cos(A – B) + cos(A + B)]
… [From (i)]
= – 1 – 2.cos C.(2.cos A.cos B)
= – 1 – 4.cos A.cos B.cos C = R.H.S.

Question 2.
sin A + sin B + sin C = 4 cos A/2 cos B/2 cos C/2
Solution:

Question 3.
cos A + Cos B + Cos C = 4 cos A/2 cos B/2 cos C/2
= \(2 \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\left(1-2 \sin ^{2} \frac{\mathrm{C}}{2}\right)\)
Solution:
L.H.S. = sin A + sin B + sin C
= \(2 \cdot \cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\left(1-2 \sin ^{2} \frac{\mathrm{C}}{2}\right)\)
In Δ ABC, A + B + C = π ,
∴ A + B = π – C

Question 4.
sin² A + sin² B – sin² C = 2 sin A sin B cos C
Solution:
We know that, sin² = \(\frac{1-\cos 2 \theta}{2}\)
L.H.S.
= sin² + sin² B + sin² C

= 1 – cos(A + B). cos(A – B) – sin²C
= (1 – sin² C ) – cos (A + B). cos (A – B)
= cos² C – cos(A + B). cos(A – B)
∴ cos(A + B) = cos(it — C)
∴ cos(A + B) = — cos C …(i)
∴ L.H.S. = cos²C + cos C.cos(A – B)
… [From (i)]
= cos C[cos C + cos(A – B)]
= cos C[- cos(A + B) + cos(A – B)]
… [From (i)]
= cos C[cos (A-B) – cos(A + B)]
= cos C(2 sin A.sin B)
= 2 sin A.sin B. cos C
= R.H.S.
[Note: The question has been modified.]

Question 5.
\(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}-\sin ^{2} \frac{C}{2}\) = \(1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\)
Solution:

Question 6.
tan \(\frac{\mathbf{A}}{2}\) tan \(\frac{\mathbf{B}}{2}\) tan \(\frac{\mathbf{B}}{2}\) tan \(\frac{\mathbf{C}}{2}\) tan \(\frac{\mathbf{C}}{2}\) tan \(\frac{\mathbf{A}}{2}\) = 1
Solution:
In Δ ABC,
A + B + C = π
∴ A + B = π – C

Question 7.
\(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}\)
Solution:
In Δ ABC,
A + B + C = π
∴ A + B = π – C

Question 8.
tan 2A + tan 2B + tan 2C = tan 2A tan 2B + tan 2C
Solution:
In Δ ABC,
A + B + C = π
∴ 2A + 2B + 2C = 2π
∴ 2A + 2B = 2π – 2C
tan(2A + 2B) = tan(2n — 2C)
\(\frac{\tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}}{1-\tan 2 \mathrm{~A} \cdot \tan 2 \mathrm{~B}}\) = -tan 2C
∴ tan2A+tan2B=—tan2C.(1-tan2A.tan2B)
∴ tan 2A + tan 2B = – tan2C+ tan2A.tan2B.tan2C
∴ tan 2A + tan 2B + tan 2C = tan2A.tan2B.tan2C

Question 9.
cos² A + cos² B – cos² C = 1 – 2 sin A sin B sin C
Solution:
we know that cos²θ = \(\frac{1+\cos 2 \theta}{2}\)
L.H.S.
= cos² A + cos² B + cos² C

= 1 + cos (A + B).cos(A — B) – cos2 C
In ΔABC,
A + B + C = π
A + B = π — C
cos(A + B) = cos(π — C)
cos(A + B) = -cosC ………….. (i)
L.H.S. = 1 — cos C.cos(A — B) — cos² C
…[From(i)]
= 1 — cos C.[cos(A — B) + cos C]
= 1 — cos C.[cos(A — B) — cos(A + B)]
.. .[From (i)]
= 1 — cos C.(2.sin A.sin B)
= 1 — 2.sinA.sin B.cos C
= R.H.S.