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

[Hint : Put $x = \cos 2\theta ]$

Putting $x = \cos \theta ,\;\;we\;get$
L.H.S. $= {\tan ^{ - 1}}\left\{ {\frac{{\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } }}} \right\}$

$= {\tan ^{ - 1}}\left\{ {\frac{{\sqrt {2{{\cos }^2}(\theta /2)} - \sqrt {2{{\sin }^2}(\theta /2)} }}{{\sqrt {2{{\cos }^2}(\theta /2)} + \sqrt {2{{\sin }^2}(\theta /2)} }}} \right\}$
[Dividing numerator and denominator by cos

$(\theta /2)]$
$= {\tan ^{ - 1}}\left\{ {\tan \left( {\frac{\pi }{4} - \frac{\theta }{2}} \right)} \right\} = \frac{\pi }{4} - \frac{\theta }{2} = \left( {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}x} \right) = R.H.S.$

Hence, L.H.S. = R.H.S.