Question

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