Given that $\tan \left\{\cos ^{-1}(x)\right\}=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
Let $\cot ^{-1} \frac{1}{2}=\phi \Rightarrow \frac{1}{2}=\cot \phi$
$\Rightarrow \sin \phi=\frac{1}{\sqrt{1+\cot ^2 \phi}}=\frac{2}{\sqrt{5}}$
$\begin{aligned}
& \text {Let } \cos ^{-1} x= \theta \Rightarrow \sec \theta=\frac{1}{x} \Rightarrow \tan \theta=\sqrt{\sec ^2 \theta-1} \\
& \Rightarrow \tan \theta=\sqrt{\frac{1}{x^2}-1} \Rightarrow \tan \theta=\frac{\sqrt{1-x^2}}{x}
\end{aligned}$
So, $\tan \left\{\cos ^{-1}(x)\right\}=\sin \left(\cot ^{-1} \frac{1}{2}\right)$
$\begin{aligned}
& \Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^2}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right) \\
& \Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \sqrt{\left(1-x^2\right) 5}=2 x
\end{aligned}$
$\Rightarrow \sqrt{5-5 x^2}=2 x$ which on squaring gives
$5-5 \mathrm{x}^2=4 \mathrm{x}^2$ Or, $x= \pm \frac{\sqrt{5}}{3}$