MCQ
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = $
  • A
    $\frac{\pi }{4}$
  • $\frac{\pi }{2}$
  • C
    $\pi $

Answer

Correct option: B.
$\frac{\pi }{2}$
b
(b) Putting $x = \tan \theta $

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

$ = {\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}$.

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