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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