Question

Prove that ${\tan }^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right]=\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x,0<x<1$

Answer 1

For this we have to put $x=\cos 2\theta$

From LHS we get,

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

Put $x=\cos 2\theta$

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

Since we know that

$\cos 2\theta =2{\cos }^2\theta -1$ or $1-2{\sin }^2\theta$

Put this value in equation 1 then we get

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

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

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

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

dividing by $\cos \theta$

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

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

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

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

By using $\tan (x+y)=\frac{\tan x-\tan y}{1+\tan x.\tan y}$

Thus, Put this value

$x=\frac{\pi }{4}$ and $y=\theta$

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

$=\frac{\pi }{4}-\theta ......2$

\since we know that

$x=\cos 2\theta$

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

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

Put this value in equation $2$ then we get

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

Hence RHS proved.