Question

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

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

Answer 1

Answer: (C)

$1$

Answer 2

Given, for $0 \leq x \leq \frac{1}{2}$
\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]
Put $\sin ^{-1} x=\theta \Rightarrow x=\sin \theta$
=\tan [\left[\sin ^{-1}\left\\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\\}-\theta\right]
=\tan \left[\sin ^{-1}\left\\{\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta\right\\}-\theta\right]
=\tan \left[\sin ^{-1}\left\\{\sin \left(\theta+\frac{\pi}{4}\right)\right\\}-\theta\right]
$=\tan \left[\theta+\frac{\pi}{4}-\theta\right]$
$=\tan \frac{\pi}{4}=1$