If ${\cos }^{-1}x={\tan }^{-1}x,$then $\sin \left({\cos }^{-1}x\right)=$

${\cos }^{-1}x={\tan }^{-1}x=\theta \left(\text{ say }\right)$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\cos \theta =\tan \theta \\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\cos }^2\theta =\sin \theta \Rightarrow {\sin }^2\theta +\sin \theta -1=0\end{array}$

$\Rightarrow \sin \theta =\frac{-1\sqrt{1+4}}{2}\Rightarrow \sin \theta =\frac{\sqrt{5}-1}{2}$

So              $x^2={\cos }^2\theta =\frac{\sqrt{5}-1}{2}$

and         $\sin \left({\cos }^{-1}x\right)=\sin \theta =\frac{\sqrt{5}-1}{2}=x^2.$