MCQ
${d \over {dx}}\left[ {{{\sin }^2}{{\cot }^{ - 1}}\left\{ {\sqrt {{{1 - x} \over {1 + x}}} } \right\}} \right]$ equals
- A$ - 1$
- ✓${1 \over 2}$
- C$ - {1 \over 2}$
- D$1$
Put $x = \cos \theta \Rightarrow \theta = {\cos ^{ - 1}}x$
==> $y = {\sin ^2}{\cot ^{ - 1}}\left\{ {\sqrt {\frac{{1 - \cos \theta }}{{1 + \cos \theta }}} } \right\} = {\sin ^2}{\cot ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)$
==> $y = {\sin ^2}\left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)$
==> $\frac{{dy}}{{d\theta }} = 2\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)\,.\,\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)\,\left( { - \frac{1}{2}} \right)$
==> $\frac{{dy}}{{d\theta }} = - \frac{{\sin (\pi - \theta )}}{2} = - \frac{{\sin \theta }}{2} = \frac{{ - 1}}{2}\,\sqrt {1 - {x^2}} $
==> $\frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\,.\,\frac{{d\theta }}{{dx}} = \frac{{ - 1}}{2}\sqrt {1 - {x^2}} \,.\,\frac{d}{{dx}}({\cos ^{ - 1}}x) = \frac{1}{2}$.
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