The value of tan-1 $\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$ $\left| x \right| < {1 \over 2},x \ne 0,$ is equal to :

Question

The value of tan^-1 $\left[ {{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} } \over {\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right],$ $\left| x \right| < {1 \over 2},x \ne 0,$ is equal to :

${\pi \over 4} + {1 \over 2}{\cos ^{ - 1}}\,{x^2}$

${\pi \over 4} + {\cos ^{ - 1}}\,{x^2}$

${\pi \over 4} - {1 \over 2}{\cos ^{ - 1}}\,{x^2}$

${\pi \over 4} - {\cos ^{ - 1}}\,{x^2}$

Solution

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