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 $
(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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