SOLUTION : $\quad \cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)=\cos \left(\cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}+\cos ^{-1} \frac{1}{5}\right)$
$=\cos \left(\frac{\pi}{2}+\cos ^{-1} \frac{1}{5}\right)=-\sin \left(\cos ^{-1}\left(\frac{1}{5}\right)\right) \\$
$=-\sqrt{1-\left(\frac{1}{5}\right)^2}=-\frac{2 \sqrt{6}}{5}$
Aliter: Let $\cos ^{-1} \frac{1}{5}=\theta \quad \Rightarrow \quad \cos \theta=\frac{1}{5}$ and $\theta \in\left(0, \frac{\pi}{2}\right)$
$\therefore \quad \sin \theta=\frac{\sqrt{24}}{5} \\$
$\therefore \quad \sin ^{-1}(\sin \theta)=\sin ^{-1}\left(\frac{\sqrt{24}}{5}\right) \\$
$\because \quad \theta \in\left(0, \frac{\pi}{2}\right) \quad \Rightarrow \quad \sin ^{-1}(\sin \theta)=\theta$
$\therefore \quad$ equation (ii) can be written as
$ \theta=\sin ^{-1}\left(\frac{\sqrt{24}}{5}\right) \quad \because \quad \theta=\cos ^{-1}\left(\frac{1}{5}\right) \\$
$\Rightarrow \quad \cos ^{-1}\left(\frac{1}{5}\right)=\sin ^{-1}\left(\frac{\sqrt{24}}{5}\right)$
Now equation (i) can be written as $y=-\sin \left\{\sin ^{-1}\left(\frac{\sqrt{24}}{5}\right)\right\}$
$\because \quad \frac{\sqrt{24}}{5} \in[-1,1] \quad \therefore \quad \sin \left\{\sin ^{-1}\left(\frac{\sqrt{24}}{5}\right)\right\}=\frac{\sqrt{24}}{5}$
$\therefore \quad$ from equation (iii), we get $y=-\frac{\sqrt{24}}{5}$
