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

MCQ

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

A
${\cot ^{ - 1}}\sqrt x $
✓
${\tan ^{ - 1}}\sqrt x $
C
${\tan ^{ - 1}}x$
D
${\cot ^{ - 1}}x$

✓

Answer

Correct option: B.

${\tan ^{ - 1}}\sqrt x $

b
(b) Let $x = {\tan ^2}\theta \Rightarrow \theta = {\tan ^{ - 1}}\sqrt x $

Now, $\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)$

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

$ = \frac{1}{2}{\cos ^{ - 1}}\cos 2\theta = \frac{{2\theta }}{2} = \theta = {\tan ^{ - 1}}\sqrt x $.

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