Question

Find the value of $\sin \left[{\cot }^{-1}\left\{\tan \left({\cos }^{-1}x\right)\right\}\right]$.

Answer 1

$\begin{array}{c}\sin \left[{\cot }^{-1}\left\{\tan \left({\cos }^{-1}x\right)\right\}\right]\\ Let{\cos }^{-1}x=t\\ \Rightarrow x=\cos t\\ \Rightarrow \cos t=x\\ \Rightarrow \sec t=\frac{1}{x}\\ 1+{\tan }^{2}t={\sec }^{2}t\\ \Rightarrow {\tan }^{2}t=\frac{1}{x^2}-1\\ \Rightarrow {\tan }^{2}t=\frac{1-x^2}{x^2}\\ \Rightarrow \tan t=\frac{\sqrt{1-x^2}}{x}\\ \Rightarrow t={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\\ So,\\ \sin \left[{\cot }^{-1}\left\{\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)\right\}\right]\\ =\sin \left[{\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right]\\ Let{\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=y\\ \Rightarrow \cot y=\frac{\sqrt{1-x^2}}{x}\\ \Rightarrow \sin y=\frac{x}{1}\\ \Rightarrow y={\sin }^{-1}x\\ \Rightarrow \sin \left[{\cot }^{-1}\frac{\sqrt{1-x^2}}{x}\right]=\sin \left({\sin }^{-1}x\right)=x\end{array}$$