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)

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