^ - 1 ( 3 + 5 x 5 + 3 x ) is equal to - Vidyadip

MCQ

${\cos ^{ - 1}}\left( {\frac{{3 + 5\cos x}}{{5 + 3\cos x}}} \right)$ is equal to

A
${\tan ^{ - 1}}\left( {\frac{1}{2}\tan \frac{x}{2}} \right)$
B
$2{\tan ^{ - 1}}\left( {2\tan \frac{x}{2}} \right)$
C
$\frac{1}{2}{\tan ^{ - 1}}\left( {2\tan \frac{x}{2}} \right)$
✓
$2{\tan ^{ - 1}}\left( {\frac{1}{2}\tan \frac{x}{2}} \right)$

✓

Answer

Correct option: D.

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

d
(d) We take $x = \frac{\pi }{2}$ then $\cos x = 0$

${\cos ^{ - 1}}\left( {\frac{{3 + 5\cos x}}{{5 + 3\cos x}}} \right) = {\cos ^{ - 1}}\left( {\frac{3}{5}} \right)$

$ = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$

Put $x = \frac{\pi }{2}$ in $2{\tan ^{ - 1}}\left( {\frac{1}{2}\tan \frac{x}{2}} \right)$

we get $2{\tan ^{ - 1}}\left( {\frac{1}{2}\tan \frac{\pi }{4}} \right)$

$ = 2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = {\tan ^{ - 1}}\left( {\frac{{2.\frac{1}{2}}}{{1 - \frac{1}{4}}}} \right)$$ = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$.

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