^ - 1 ( 1 + x^2 - 1 x ) =

MCQ

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

A
${\tan ^{ - 1}}x$
✓
$\frac{1}{2}{\tan ^{ - 1}}x$
C
$2{\tan ^{ - 1}}x$
D
None of these

✓

Answer

Correct option: B.

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

b
(b) ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right]$

(Putting $x = \tan \theta )$

$ = {\tan ^{ - 1}}\left[ {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right] = {\tan ^{ - 1}}\left[ {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right]$

$ = {\tan ^{ - 1}}\left[ {\frac{{2\,{{\sin }^2}\frac{\theta }{2}}}{{2\,\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right]$

$ = {\tan ^{ - 1}} (\tan \frac{\theta }{2} )= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$.

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