Prove that ${\tan ^{ – 1}}\left( {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 – {x^2}} }}{{\sqrt {1 + {x^2}} – \sqrt {1 – {x^2}} }}} \right) = \frac{\pi }{4} + \frac{1}{2}{\cos ^{ – 1}}{x^2}$.
[NCERT,Exemplar.2.3,Q.12,Page.36]

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.

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