Prove the following identity. 1 + cos θ 1 - cos θ = ( cosec θ + cot θ ) 2

LHS: 1 + cos θ 1 - cos θ Rationalizing the denominator, we get 1 + cos θ 1 - cos θ × 1 + cos θ 1 + cos θ Applying ( a - b ) ( a + b ) = a 2 - b 2 in the denominator, we get ⇒ 1 + cos θ 2 1 - cos 2 θ We know that, 1 - cos 2 θ = sin 2 θ = 1 + cos θ sin θ 2 = 1 sin θ + cos θ sin θ 2 We know that, 1 sin θ = cosec θ and cos θ sin θ = cot θ Therefore, 1 sin θ + cos θ sin θ 2 = cosec θ + cot θ 2 = RHS

Prove the following identities 1 + tan α · tan β 2 + tan α - tan β 2 = sec 2 α · sec 2 β

Consider L.H.S = 1 + tan α tan β 2 + tan α – tan β 2 = 1 + tan 2 α tan 2 β + 2 tan α tan β + tan 2 α + tan 2 β – 2 tan α tan β = 1 + tan 2 α + tan 2 α tan 2 β + tan 2 β = 1 + tan 2 α + tan 2 β ( 1 + tan 2 α ) = ( 1 + tan 2 α ) ( 1 + tan 2 β ) = sec 2 α sec 2 β = R.H.S

If cos θ + sin θ = 2 cos θ , prove that cos θ - sin θ = 2 sin θ .

Given, cos θ + sin θ = 2 cos θ Transposing cos θ to RHS, we get, ⇒ sin θ = 2 cos θ - cos θ ⇒ sin θ = cos θ 2 - 1 ⇒ sin θ 2 - 1 = cos θ ⇒ sin θ 2 - 1 × 2 + 1 2 + 1 = cos θ ⇒ 2 + 1 sin θ 2 2 - 1 2 = cos θ ⇒ 2 + 1 sin θ = cos θ ⇒ 2 sin θ + sin θ = cos θ ⇒ 2 sin θ + sin θ = cos θ ⇒ cos θ - sin θ = 2 sin θ Hence, proved.

If tan θ = a b , then show that a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 .

Given that, tan θ = a b . We have to prove that, a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 LHS = a sin θ - b cos θ a sin θ + b cos θ = a sin θ cos θ - b cos θ cos θ a sin θ cos θ + b cos θ cos θ [Divide both the numerator and the denominator by cos θ ] = a tan θ - b a tan θ + b = a × a b - b a × a b + b [Substitute the value of tan θ = a b that is given] = a 2 - b 2 a a 2 + b 2 b = a 2 - b 2 a 2 + b 2 = RHS [Proved]

If sin x + cos x = m , then prove that sin 6 x + cos 6 x = 4 - 3 m 2 - 1 2 4 , where m 2 ≤ 2 .

Given, m = sin x + cos x ∴ m 2 = sin 2 x + cos 2 x + 2 sin x cos x = 1 + 2 sin x cos x ⇒ m 2 – 1 = 2 sin x cos x Now, sin 6 x + cos 6 x = sin 2 x 3 + cos 2 x 3 = sin 2 x + cos 2 x 3 – 3 sin 2 x cos 2 x ( sin 2 x + cos 2 x ) = 1 – 3 sin 2 x cos 2 x = 1 – 3 m 2 – 1 2 2 = 4 – 3 m 2 – 1 2 4 Hence, proved.

If x sin 3 θ + y cos 3 θ = sin θ · cos θ and x sin θ - y cos θ = 0 , prove that x 2 + y 2 = 1 .

x sin 3 θ + y cos 3 θ = sin θ cos θ . .. ( 1 ) and x sin θ − y cos θ = 0 . .. ( 2 ) From 2 x sin θ = y cos θ Multiple both sides by sin 2 θ x sin 3 θ = y cos θ sin 2 θ Put in 1 y cos θ sin 2 θ + y cos 3 θ = sin θ cos θ y cos θ ( sin 2 θ + cos 2 θ ) = sin θ . cos θ , using sin 2 θ + cos 2 θ = 1 y = sin θ Similarly x = cos θ Square and add to get the result. x 2 + y 2 = 1

E M B I B E

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 2 - Trigonometric Functions and Their Graphs>EXERCISE 2.1

K. C. Sinha Solutions for Chapter: Trigonometric Functions and Their Graphs, Exercise 1: EXERCISE 2.1

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Trigonometric Functions and Their Graphs, Exercise 1: EXERCISE 2.1

Attempt the practice questions on Chapter 2: Trigonometric Functions and Their Graphs, Exercise 1: EXERCISE 2.1 with hints and solutions to strengthen your understanding. TRIGONOMETRY FOR JEE MAIN AND ADVANCED solutions are prepared by Experienced Embibe Experts.