Question

The value of ${\tan }^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ is equal to

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

Answer 1

Answer: (C)

We have, ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$
Let $x^2=\cos 2\theta$. Then, we get
Given expression $={\tan }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}\right)$
$={\tan }^{-1}\left(\frac{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta }\right)$
$\left[\because 1+\cos 2\theta =2{\cos }^2\theta \text{ and }1-\cos 2\theta =2{\sin }^2\theta \right]$
$={\tan }^{-1}\left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)$
(divide numerator and denominator inside the bracket by $\cos \theta$ )
$={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)$
$={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\left(\tan \frac{\pi }{4}\right)(\tan \theta )}\right)={\tan }^{-1}\left[\tan \left(\frac{\pi }{4}+\theta \right)\right]$
$\left(\because \tan (\theta +\varphi )=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }\right)$
$=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$
$\left(\because x^2=\cos 2\theta \Rightarrow \theta =\frac{1}{2}{\cos }^{-1}{x}^2\right)$