MCQ
The value of $\sin {\cot ^{ - 1}}\tan {\cos ^{ - 1}}x$ is equal to
- ✓$x$
- B$\frac{x }{2}$
- C$2x$
- DNone of these
$ \Rightarrow \,\,\tan \theta = \sqrt {{{\sec }^2}\theta - 1} = \sqrt {\frac{1}{{{x^2}}} - 1} = \frac{1}{x}\sqrt {1 - {x^2}} $
Now $\sin \,\,{\cot ^{ - 1}}\tan \theta = \sin \,\,{\cot ^{ - 1}}\,\left( {\frac{1}{x}\sqrt {1 - {x^2}} } \right)$
Again, putting $x = \sin \theta $
$\sin \,\,{\cot ^{ - 1}}\left( {\frac{1}{x}\sqrt {1 - {x^2}} } \right) = \sin \,\,{\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \right)$
$ = \sin \,\,{\cot ^{ - 1}}\,(\cot \theta ) = \sin \theta = x$.
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