In △ A B C , prove that sin B - C 2 = b - c a cos A 2 .

Consider the given equation sin B - C 2 = b - c a cos A 2 . We know that by Sine rule ⇒ a sin A = b sin B = c sin C = k ∴ ⇒ a = k sin A , b = k sin B and c = k sin C Now take the right hand side of the given equation: R . H . S . = b - c a cos A 2 = k sin B - sin C k sin A cos A 2 = sin B - sin C sin A cos A 2 ∵ sin C - sin D = 2 cos C + D 2 sin C - D 2 , sin 2 A = 2 sin A cos A ⇒ R . H . S . = 2 cos B + C 2 sin B - C 2 2 sin A 2 cos A 2 cos A 2 = cos π 2 - A 2 sin B - C 2 sin A 2 ∵ In △ ABC , A + B + C = π = sin A 2 sin B - C 2 sin A 2 = sin B - C 2 ⇒ R . H . S . = L . H . S . Hence proved.

Show that sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16

We have to show: sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16 Let, sin - 1 5 13 = x ⇒ sin x = 5 13 And cos 2 x = 1 - sin 2 x = 1 - 25 169 = 144 169 ⇒ cos x = 144 169 = 12 13 ∴ tan x = sin x cos x = 5 13 12 13 = 5 12 Again, let cos - 1 3 5 = y ⇒ cos y = 3 5 ∴ sin y = 1 - cos 2 y = 1 - 3 5 2 = 1 - 9 25 ⇒ sin y = 16 25 = 4 5 ⇒ tan y = sin y cos y = 4 5 3 5 = 4 3 We know that, tan ( x + y ) = tan x + tan y 1 - tan x tan y ∴ tan ( x + y ) = 5 12 + 4 3 1 - 5 12 4 3 ⇒ tan ( x + y ) = 5 + 16 12 36 - 20 36 ⇒ tan ( x + y ) = 21 × 3 16 ⇒ tan ( x + y ) = 63 16 ⇒ x + y = tan - 1 63 16 ⇒ sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16 Hence, proved.

In △ A B C , with usual notations prove that a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = c 2 .

Consider a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = c 2 Taking L . H . S . = a - b 2 cos 2 C 2 + a + b 2 sin 2 C 2 = a 2 + b 2 - 2 a b cos 2 C 2 + a 2 + b 2 + 2 a b sin 2 C 2 = a 2 + b 2 cos 2 C 2 + sin 2 C 2 + 2 a b sin 2 C 2 - cos 2 C 2 = a 2 + b 2 - 2 a b cos 2 C 2 - sin 2 C 2 = a 2 + b 2 - 2 a b cos C We know that, cos C = a 2 + b 2 - c 2 2 a b L . H . S . = a 2 + b 2 - 2 a b a 2 + b 2 - c 2 2 a b = a 2 + b 2 - a 2 + b 2 - c 2 = a 2 + b 2 - a 2 - b 2 + c 2 = c 2 L . H . S . = R . H . S . Hence, proved.

In △ A B C , if a + b + c = 2 s , then prove that sin ⁡ A 2 = ( s − b ) ( s − c ) b c , with usual notations.

Given: a + b + c = 2 s To prove: sin ⁡ A 2 = ( s − b ) ( s − c ) b c We know that △ A B C = 1 2 · b c · sin ⁡ A . . . 1 and also using herons formula Δ A B C = s ( s − a ) ( s − b ) ( s − c ) . . . 2 Using 1 and 2 , we get: 1 2 · b c · sin ⁡ A = s ( s − a ) ( s − b ) ( s − c ) ⇒ sin ⁡ A = 2 s ( s − a ) ( s − b ) ( s − c ) b c ⇒ 2 sin ⁡ A 2 cos ⁡ A 2 = 2 s ( s − a ) ( s − b ) ( s − c ) b c ∵ sin 2 θ = 2 sin θ cos θ ⇒ sin ⁡ A 2 cos ⁡ A 2 = s ( s − a ) ( s − b ) ( s − c ) b c . . . * Now, we know that cos ⁡ A = b 2 + c 2 − a 2 2 b c [using cosine rule] and cos ⁡ 2 A = 2 cos 2 ⁡ A − 1 . ⇒ 2 cos 2 ⁡ A 2 − 1 = b 2 + c 2 − a 2 2 b c ⇒ 2 cos 2 ⁡ A 2 = b 2 + c 2 − a 2 2 b c + 1 ⇒ 2 cos 2 ⁡ A 2 = b 2 + c 2 − a 2 + 2 b c 2 b c ⇒ cos 2 ⁡ A 2 = b 2 + c 2 − a 2 + 2 b c 4 b c ⇒ cos 2 ⁡ A 2 = ( b + c − a ) ( b + c + a ) 4 b c ⇒ cos 2 ⁡ A 2 = ( 2 s - a − a ) ( 2 s ) 4 b c ∵ b + c + a = 2 s ⇒ cos 2 ⁡ A 2 = ( s - a ) ( 4 s ) 4 b c ⇒ cos 2 ⁡ A 2 = s ( s - a ) b c ⇒ cos ⁡ A 2 = s ( s - a ) b c . . . 3 Putting the equation 3 in equation * , we get: sin ⁡ A 2 s ( s − a ) b c = s ( s − a ) ( s − b ) ( s − c ) b c ⇒ sin ⁡ A 2 = ( s − b ) ( s − c ) b c Hence proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 3 - Trigonometric Functions>Maharastra Board-2018

Embibe Experts Solutions for Chapter: Trigonometric Functions, Exercise 1: Maharastra Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Trigonometric Functions, Exercise 1: Maharastra Board-2018

Attempt the free practice questions on Chapter 3: Trigonometric Functions, Exercise 1: Maharastra Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.