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

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

LHS=$\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )........(i)$

$\left [ x^{2}=\cos2\theta=\cos^{2}\theta+\sin^{2}\theta=1-2\sin^{2}\theta = 2\cos^{2}\theta-1 \right ]$

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

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

$\therefore \sqrt{1+x^{2}}=\sqrt{1+\cos2\theta}$

$\Rightarrow \sqrt{1+2\cos^{2}\theta-1}=\sqrt{2}\cos \theta$

And  $\sqrt{1-x^{2}}=\sqrt{1-\cos2\theta}$

$\sqrt{1-1+2 \sin^{2}\theta}=\sqrt{2}\sin \theta$

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

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

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

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

$= \tan^{-1}\left [ \tan\left ( \frac{\pi}{4}+\theta \right ) \right ]$  $\left [ Since , \tan\left ( x+y \right )=\frac{\tan x+\tan y}{1-\tan x.\tan y} \right ]$

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

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

=RHS

LHS=RHS

Hence Proved