Question

Prove ${\tan }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi }{4}-\frac{1}{2}\cos {}^{-1}x$

Answer 1

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

$puttingx=\cos 2\theta$

${\cos }^{-1}x=2\theta$

$\theta =\frac{1}{2}{\cos }^{-1}x$

$={\tan }^{-1}\left(\frac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos \theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos \theta }}\right)$

$={\tan }^{-1}\left(\frac{\sqrt{2{\cos }^2\theta }-\sqrt{2{\sin }^2\theta }}{\sqrt{2{\cos }^2\theta }+\sqrt{2{\sin }^2\theta }}\right)$

$={\tan }^{-1}\left(\frac{\cos \theta -sin\theta }{\cos \theta +sin\theta }\right)$

$={\tan }^{-1}\left(\frac{1-\tan \theta }{1+\tan \theta }\right)$

$={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }\right)$

$={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\theta \right)\right)$

$=\frac{\pi }{4}-\theta$

$=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$

LHS=RHS

Hence proof