Solve for x : tan - 1 x - 1 x - 2 + tan - 1 x + 1 x + 2 = π 4

Given that, tan - 1 x - 1 x - 2 + tan - 1 x + 1 x + 2 = π 4 ⇒ tan - 1 x - 1 x - 2 + x + 1 x + 2 1 - x - 1 x - 2 × x + 1 x + 2 = π 4 [Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x y < 1 ] ⇒ tan - 1 x 2 + 2 x - x - 2 + x 2 + x - 2 x - 2 x 2 - 4 - x 2 + 1 = π 4 ⇒ tan - 1 x 2 - 4 + x 2 - 3 = π 4 ⇒ x 2 - 4 + x 2 - 3 = tan π 4 ⇒ 2 x 2 - 4 - 3 = 1 ⇒ 2 x 2 = - 3 + 4 ⇒ 2 x 2 = 1 ⇒ x = ± 1 2 Thus, the required solution x = ± 1 2 .

If sec - 1 x = cosec - 1 y , show that 1 x 2 + 1 y 2 = 1 .

Given that, sec - 1 x = cosec - 1 y and to prove: 1 x 2 + 1 y 2 = 1 Therefore, cos - 1 1 x = sin - 1 1 y ⇒ sin - 1 1 - 1 x 2 = sin - 1 1 y ⇒ 1 - 1 x 2 = 1 y Squaring on both sides, we have: 1 - 1 x 2 = 1 y 2 ⇒ 1 = 1 x 2 + 1 y 2 Hence, proved.

If cos - 1 x 2 + cos - 1 y 3 = θ , then prove that: 9 x 2 - 12 x y cos θ + 4 y 2 = 36 sin 2 θ .

Given that, cos - 1 x 2 + cos - 1 y 3 = θ and To prove: 9 x 2 - 12 x y cos θ + 4 y 2 = 36 sin 2 θ Consider, cos - 1 x 2 + cos - 1 y 3 = θ ⇒ cos - 1 x 2 × y 3 - 1 - x 2 2 1 - y 3 2 = θ [Since cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 1 - y 2 ] ⇒ x 2 × y 3 - 1 - x 2 2 1 - y 3 2 = cos θ ⇒ x y 6 - 1 - x 2 2 2 1 - y 2 3 2 = cos θ ⇒ x y 6 - cos θ = 1 - x 2 4 1 - y 2 9 ⇒ x y 6 - cos θ 2 = 1 - x 2 4 1 - y 2 9 2 ⇒ x y 6 2 + cos 2 θ - 2 × x y 6 cos θ = 1 - x 2 4 1 - y 2 9 ⇒ x y 6 2 + cos 2 θ - x y 3 cos θ = 1 - y 2 9 - x 2 4 + x y 6 2 ⇒ y 2 9 + x 2 4 - x y 3 cos θ = 1 - cos 2 θ ⇒ y 2 9 + x 2 4 - x y 3 cos θ = sin 2 θ ∵ cos 2 x + sin 2 x = 1 , x ∈ R ⇒ 9 x 2 - 12 x y cos θ + 4 y 2 = 36 sin 2 θ [Multiplied both sides with 36 ] Hence, proved.

Evaluate cos 2 cos - 1 x + sin - 1 x at x = 1 5 .

We have, cos 2 cos - 1 x + sin - 1 x = cos cos - 1 x + cos - 1 x + sin - 1 x = cos cos - 1 x + π 2 ∵ cos - 1 A + sin - 1 A = π 2 , ∀ A ∈ [ - 1 , 1 ] = - sin cos - 1 x ∵ cos π 2 + A = - sin A = - sin sin - 1 1 - x 2 = - 1 - x 2 ∵ sin sin - 1 A = A = - 1 - 1 5 2 ∵ x = 1 5 = - 1 - 1 25 = - 24 25 = - 24 5

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 2 - Inverse Trigonometric Functions>ICSE-2019

Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: ICSE-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Inverse Trigonometric Functions, Exercise 1: ICSE-2018

Attempt the free practice questions on Chapter 2: Inverse Trigonometric Functions, Exercise 1: ICSE-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.