Prove $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1$ [Hint: put $x=\cos 2 \theta$ ]
Put $x=\cos 2 \theta$ so that $\theta=\frac{1}{2} \cos ^{-1} x$. Then, we have:
L.H.S. $=\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$
$=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right)$
$=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right)$
$=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right)$
$=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right)=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)$
$=\tan ^{-1} 1-\tan ^{-1}(\tan \theta)$ $\left[\tan ^{-1}\left(\frac{x-y}{1+x y}\right)=\tan ^{-1} x-\tan ^{-1} y\right]$
$=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x=$ R.H.S $.$
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