SOLUTION : LHS $=\tan ^{-1}\left[\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right]$
Let $\cos ^{-1} x=\theta$, so that $x=\cos \theta$ and $0 \leq \theta \leq \frac{3 \pi}{4}$
$\therefore \quad \tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right)=\tan ^{-1}\left(\frac{\sqrt{2} \cos \frac{\theta}{2}-\sqrt{2} \sin \frac{\theta}{2}}{\sqrt{2} \cos \frac{\theta}{2}+\sqrt{2} \sin \frac{\theta}{2}}\right] \\$
$\left(\because 1+\cos \theta=2 \cos ^2(\theta / 2) \text { and } 1-\cos \theta=2 \sin ^2(\theta / 2)\right)$
$=\tan ^{-1}\left[\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}\right]$
inside the bracket divide numerator and denomerator by $\cos \frac{\theta}{2}$.
$=\tan ^{-1}\left[\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}\right]=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right]=\frac{\pi}{4}-\frac{\theta}{2}\left(0 \leq \theta \leq \frac{3 \pi}{4} \Rightarrow \frac{\pi}{4} \geq \frac{\pi}{4}-\frac{\theta}{2} \geq-\frac{\pi}{4}\right) \\$
$=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x=\text { RHS }$
