Given, tan -1 $\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$ Let x 2 = cos $\theta$ = tan -1 $\left[ {{{\sqrt {1 + \cos \theta } + \sqrt {1 - \cos \theta } } \over {\sqrt {1 + \cos \theta } - \sqrt {1 - \cos \theta } }}} \right]$ = tan -1 $\left[ {{{\sqrt {2{{\cos }^2}{\theta \over 2}} + \sqrt {2{{\sin }^2}{\theta \over 2}} } \over {\sqrt {2{{\cos }^2}{\theta \over 2}} - \sqrt {2{{\sin }^2}{\theta \over 2}} }}} \right]$ = tan -1 $\left[ {{{\sqrt 2 \cos {\theta \over 2} + \sqrt 2 \sin {\theta \over 2}} \over {\sqrt 2 \cos {\theta \over 2} - \sqrt 2 \sin {\theta \over 2}}}} \right]$ = tan -1 $\left[ {{{\cos {\theta \over 2} + \sin {\theta \over 2}} \over {\cos {\theta \over 2} - \sin {\theta \over 2}}}} \right]$ = tan -1 $\left[ {{{1 + \tan {\theta \over 2}} \over {1 - \tan {\theta \over 2}}}} \right]$ = tan -1 $\left[ {{{\tan {\pi \over 4} + \tan {\theta \over 2}} \over {1 - \tan {\pi \over 4} + \tan {\theta \over 2}}}} \right]$ = tan -1 $\left( {\tan \left( {{\pi \over 4} + {\theta \over 2}} \right)} \right)$ = ${\pi \over 4} + {\theta \over 2}$ = ${\pi \over 4} + {1 \over 2}$ cos -1 x 2