Prove the following: 1 tan 3 A - tan A - 1 cot 3 A - cot A = cot 2 A

LHS = 1 tan 3 A - tan A - 1 tan 3 A - tan A = 1 sin 3 A cos 3 A - sin A cos A - 1 cos 3 A sin 3 A - cos A sin A = cos 3 A cos A sin 3 A cos A - sin A cos A - sin 3 A sin A cos 3 A sin A - cos A sin 3 A = cos 3 A cos A sin 3 A - A - sin 3 A sin A sin A - 3 A ∵ sin ( x - y ) = sin x cos y - cos x sin y = cos 3 A cos A sin 2 A + sin 3 A sin A sin 2 A = cos 3 A cos A + sin 3 A sin A sin 2 A = cos 3 A - A sin 2 A ∵ cos x cos y + sin x sin y = cos x - y = cos 2 A sin 2 A = cot 2 A = RHS Hence, proved.

Prove the following: cos 7 ° cos 14 ° cos 28 ° cos 56 ° = sin 68 ° 16 cos 83 °

LHS = cos 7 ° cos 14 ° cos 28 ° cos 56 ° Multiply and divide with 2 sin 7 ° . = 1 2 sin 7 ° 2 sin 7 ° cos 7 ° · cos 14 · cos 28 · cos 56 ° = 1 2 sin 7 ° sin 14 ° cos 14 ° cos 28 cos 56 ° ∵ 2 sin A cos A = sin 2 A = 1 2 × 1 2 sin 7 ° sin 28 ° cos 28 ° cos 6 ° = 1 2 × 1 4 sin 7 ° sin 56 ° cos 56 ° = 1 2 × 1 8 sin 7 ° sin 112 ° = 1 16 × sin 180 ° - 68 ° sin 90 ° - 83 ° = sin 68 ° 16 cos 83 ° ∵ sin 180 ° - θ = sinθ & sin 90 ° - θ = cosθ = RHS Hence, proved.

Prove the following: sin 2 - 160 ° sin 2 70 ° + sin 180 ° - θ sinθ = sec 2 20 °

LHS = sin 2 - 160 ° sin 2 70 ° + sin ( 180 ° - θ ) sinθ = - sin ( 160 ° ) 2 sin 2 70 ° + sinθ sinθ ∵ sin ( 180 - θ ) = sinθ & sin - θ = - sinθ = sin ( 90 ° + 70 ° ) 2 sin 2 70 ° + 1 = cos 2 70 ° sin 2 70 ° + 1 ∵ sin ( 90 ° + θ ) = cosθ = cot 2 70 ° + 1 = cosec 2 70 ° ∵ 1 + cot 2 θ = cosec 2 θ = cosec 2 ( 90 ° - 70 ° ) = sec 2 20 ° ∵ cosec ( 90 ° - θ ) = sec θ = RHS Hence, proved.

Prove the following: 2 cos 4 x + 1 2 cos x + 1 = 2 cos x - 1 2 cos 2 x - 1

LHS = 2 cos 4 x + 1 2 cos x + 1 = 2 cos 2 2 x + 1 2 cos x + 1 = 2 2 cos 2 2 x - 1 + 1 2 cos x + 1 ∵ cos 2 A = 2 cos 2 A - 1 = 4 cos 2 2 x - 1 2 cos x + 1 = 2 cos 2 x - 1 2 cos 2 x + 1 2 cos x + 1 = 2 2 cos 2 x - 1 + 1 2 cos 2 x - 1 2 cos x + 1 ∵ cos 2 A = 2 cos 2 A - 1 = 4 cos 2 x - 1 2 cos 2 x - 1 2 cos x + 1 = 2 cos x - 1 2 cos x + 1 2 cos 2 x - 1 2 cos x + 1 ∵ a 2 - b 2 = a - b a + b = 2 cos x - 1 2 cos 2 x - 1 = RHS Hence, proved.

Prove the following: cos 2 x + cos 2 ( x + 120 ° ) + cos 2 ( x - 120 ° ) = 3 2

LHS = cos 2 x + cos 2 ( x + 120 ° ) + cos 2 ( x - 120 ° ) = 1 + cos 2 x 2 + 1 + cos 2 x + 120 ° 2 + 1 + cos 2 x - 120 ° 2 ∵ cos 2 A = 2 cos 2 A - 1 ⇒ cos 2 A = 1 + cos 2 A 2 = 1 2 3 + cos 2 x + cos ( 2 x + 240 ° ) + cos ( 2 x - 240 ° ) = 1 2 3 + cos 2 x + 2 cos 2 x cos 240 ° ∵ cos A + cos B = 2 cos A + B 2 cos A - B 2 = 1 2 3 + cos 2 x + 2 cos 2 x cos 180 + 60 ° = 1 2 3 + cos 2 x + 2 cos 2 x - cos 60 ° = 1 2 3 + cos 2 x + 2 cos 2 x · - 1 2 = 1 2 [ 3 + cos 2 x - cos 2 x ] = 3 2 = RHS Hence, proved.

Prove the following: 2 cosec 2 x + cosec x = sec x cot x 2

LHS = 2 cosec 2 x + cosec x = 2 sin 2 x + 1 sin x = 2 2 sin x cos x + 1 sin x ∵ sin 2 A = 2 sin A cos A = 1 sin x cos x + 1 sin x = 1 + cos x sin x cos x = 1 + cos x sin x · sec x = 2 cos 2 x 2 2 sin x 2 cos x 2 · sec x ∵ 1 + cos 2 A = 2 cos 2 A = cos x 2 sin x 2 · sec x = sec x · cot x 2 = RHS Hence, proved.

Prove the following: 4 cos x cos x + π 3 cos x - π 3 = cos 3 x

To prove, 4 cos x cos x + π 3 cos x - π 3 = cos 3 x LHS = 4 cos x · cos x + π 3 · cos π - π 3 = 4 cos x · cos x cos π 3 - sin x sin π 3 · cos x cos π 3 + sin x sin π 3 ∵ cos ( A ± B ) = cos A cos B ∓ sin A sin B = 4 cos x · 1 2 cos x - 3 2 sin x · 1 2 cos x + 3 2 sin x = 4 cos x · 1 4 cos 2 x - 3 4 sin 2 x = cos 3 x - 3 sin 2 x cos x = cos 3 x - 3 1 - cos 2 x cos x = cos 3 x - 3 cos x + 3 cos 3 x = cos 3 x = RHS

Prove the following: sin x tan x 2 + 2 cos x = 2 1 + tan 2 x 2

LHS = sin x tan x 2 + 2 cos x = 2 tan x 2 1 + tan 2 x 2 · tan x 2 + 2 1 - tan 2 x 2 1 + tan 2 x 2 ∵ sin A = 2 tan A 2 1 + tan 2 A 2 & cos A = 1 - tan 2 A 2 1 + tan 2 A 2 = 2 tan 2 x 2 + 2 - 2 tan 2 x 2 1 + tan 2 x 2 = 2 1 + tan 2 x 2 = RHS Hence, proved.

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Mathematics and Statistics (Arts & Science) Part 1>Chapter 3 - Trigonometry - II>EXERCISE 3.3

Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 3: EXERCISE 3.3

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Trigonometry - II, Exercise 3: EXERCISE 3.3

Attempt the practice questions on Chapter 3: Trigonometry - II, Exercise 3: EXERCISE 3.3 with hints and solutions to strengthen your understanding. Mathematics and Statistics (Arts & Science) Part 1 solutions are prepared by Experienced Embibe Experts.