The value of $$\sin {\cot ^{ - 1}}\tan {\cos ^{ - 1}}x,{\text{ is}}$$

Solution :
$$\eqalign{ & {\text{Let, }}{\cos ^{ - 1}}x = \theta \cr & \Rightarrow x = \cos \theta {\text{ or }}\sec \theta = \frac{1}{x} \cr & \Rightarrow \tan \theta = \sqrt {{{\sec }^2}\theta - 1} = \sqrt {\frac{1}{{{x^2}}} - 1} = \frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} \cr & {\text{Now, }}\sin {\cot ^{ - 1}}\tan \theta = \sin {\cot ^{ - 1}}\left( {\frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} } \right). \cr} $$
Again, putting $$x = \sin \theta ,$$   we get
$$\eqalign{ & \sin {\cot ^{ - 1}}\left( {\frac{1}{{\left| x \right|}}\sqrt {1 - {x^2}} } \right) = \sin {\cot ^{ - 1}}\left( {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \right) \cr & = \sin {\cot ^{ - 1}}\left| {\cot \theta } \right| = \sin \theta = x. \cr} $$