The value of ^ - 1 [ 1 + x ^ 2 + 1 - x ^ 2 1 + x ^ 2 - 1 - x ^ 2 ], |x| < (1)/(2), x ≠ 0, is equal to:

Correct answer: A. (π)/(4) + (1)/(2) ^ - 1 x ^ 2 Assume that , x ^ 2 = 2θ; θ = (1)/(2) ^ -1 x ^ 2 ∴ ^ - 1 [ 1 + 2θ + 1 - 2θ 1 + 2θ - 1 - 2θ ] = ^ - 1 [ 2 ^ 2 θ + 2 ^ 2 θ 2 ^ 2 θ - 2 ^ 2 θ ] = ^ - 1 [(cosθ + sinθ)/(cosθ - sinθ)] = ^ - 1 [(1 + tan θ)/(1 - tan θ)] = ^ - 1 [ ((π)/(4) + θ)] = (π)/(4) + (1)/(2) ^ - 1 x ^ 2

JEE Main 20178 AprilOnline

Exam: JEE Main
Year: 2017
Date: 8 April
Mode: Online
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2017 · 8 April · Online · Mathematics · Inverse Trigonometric Functions

The value of ^ - 1 [ 1 + x ^ 2 + 1 - x ^ 2 1 + x ^ 2 - 1 - x ^ 2 ],...

The value of

\tan^{- 1} \left[\dfrac{\sqrt{1 + {x}^{2}} + \sqrt{1 - {x}^{2}}}{\sqrt{1 + {x}^{2}} - \sqrt{1 - {x}^{2}}}\right], \left|x\right| < \dfrac{1}{2}, x \ne 0,

is equal to:

Options

A $\dfrac{\pi}{4} + \dfrac{1}{2}\cos^{- 1} {x}^{2}$
B $\dfrac{\pi}{4} - \cos^{- 1} {x}^{2}$
C $\dfrac{\pi}{4} - \dfrac{1}{2}\cos^{- 1} {x}^{2}$
D $\dfrac{\pi}{4} + \cos^{- 1} {x}^{2}$

Correct answer

A. $\dfrac{\pi}{4} + \dfrac{1}{2}\cos^{- 1} {x}^{2}$

Step-by-step solution

Assume that ,

{x}^{2} = \cos 2\theta; \theta = \dfrac{1}{2}\cos^{-1}{x}^{2}

∴ \tan^{- 1} \left[\dfrac{\sqrt{1 + \cos 2\theta} + \sqrt{1 - \cos 2\theta}}{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}}\right] = \tan^{- 1} \left[\dfrac{\sqrt{2\cos^{2}\theta} + \sqrt{2\sin^{2}\theta}}{\sqrt{2\cos^{2}\theta} - \sqrt{2\sin^{2}\theta}}\right] = \tan^{- 1} \left[\dfrac{cos\theta + sin\theta}{cos\theta - sin\theta}\right]

= \tan^{- 1} \left[\dfrac{1 + tan \theta}{1 - tan \theta}\right]

= \tan^{- 1} \left[\tan \left(\dfrac{\pi}{4} + \theta\right)\right]

= \dfrac{\pi}{4} + \dfrac{1}{2}\cos^{- 1} {x}^{2}

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