Given 0 ≤ x ≤ (1)/(2) , then the value of ( ^ -1 ( x 2 + 1-x^2 2 )- ^ -1 x) is

Correct answer: A. 1 aligned & [ ^ -1 ( x 2 + 1-x^2 2 )- ^ -1 x] & = [ ^ -1 ( x+ 1-x^2 2 )- ^ -1 x] & = [ ^ -1 ( θ+ θ 2 )-θ] [ array l Put ^ -1 x=θ x= θ array ] & = [ ^ -1 [ (θ+(π)/(4))]-θ] & = (θ+(π)/(4)-θ) & = (π)/(4)=1 aligned

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

Given 0 ≤ x ≤ (1)/(2) , then the value of ( ^ -1 ( x 2 + 1-x^2 2 )-...

Given

0 \leq x \leq \frac{1}{2}

, then the value of

\tan \left(\sin ^{-1} \left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1}

x\right) is

Options

A $1$
B $\sqrt 3$
C $-1$
D $\frac{1}{\sqrt{3}}$

Correct answer

A. $1$

Step-by-step solution

\begin{aligned} & \tan \left[\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right] \\ & =\tan \left[\sin ^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right] \\ & =\tan \left[\sin ^{-1}\left(\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right)-\theta\right] \quad \ldots\left[\begin{array}{l}\text { Put } \sin ^{-1} x=\theta \\ \Rightarrow x=\sin \theta\end{array}\right] \\ & =\tan \left[\sin ^{-1}\left[\sin \left(\theta+\frac{\pi}{4}\right)\right]-\theta\right] \\ & =\tan \left(\theta+\frac{\pi}{4}-\theta\right) \\ & =\tan \frac{\pi}{4}=1\end{aligned}

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