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

$therefore,$ ${\rm{LHS}} = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right)$

….(i)
[let ${x^2} = \cos 2\theta = \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = 1 - 2{\sin ^2}\theta = 2{\cos ^2}\theta - 1$]
$\Rightarrow$ ${\cos ^{ - 1}}{x^2} = 2\theta \Rightarrow \theta = \frac{1}{2}{\cos ^{ - 1}}{x^2}$

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

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

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

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

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

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

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

$= \frac{\pi }{4} + \theta = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ - 1}}{x^2}$
$= {\rm{RHS}}$ Hence proved.