STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}$

$ = {\tan ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\cot 2\theta ) + {\cos ^{ - 1}}(\cos 2\theta )$

$ = \frac{\pi }{2} - 2\theta + 2\theta = \frac{\pi }{2}$.