MCQ
${d \over {dx}}\left( {{{\tan }^{ - 1}}{{\cos x} \over {1 + \sin x}}} \right) = $
- ✓$ - {1 \over 2}$
- B${1 \over 2}$
- C$ - 1$
- D$1$
$= \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos \frac{x}{2}}}} \right)} \right]$
$ = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{1 - \tan \left( {\frac{x}{2}} \right)}}{{1 + \tan \left( {\frac{x}{2}} \right)}}} \right)} \right] = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\tan \left( {\frac{\pi }{4} - \frac{x}{2}} \right)} \right] = - \frac{1}{2}$
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