PYQ NEET- Inverse Trigonometric Functions

• 2016:

The answer is $\pi$.

We know that $\sin^{-1}(2\sin^{-1}x)=2\sin^{-1}x-\frac{\pi}{2}$. So, $\sin^{-1}(2\sin^{-1}x)+\cos^{-1}(\cos 2x)=2\sin^{-1}x-\frac{\pi}{2}+\cos^{-1}(1-2\sin^2x)$. We also know that $\cos^{-1}(1-2\sin^2x)=\sin^{-1}(2\sin x)$. So, $\sin^{-1}(2\sin^{-1}x)+\cos^{-1}(\cos 2x)=2\sin^{-1}x-\frac{\pi}{2}+2\sin^{-1}(\sin x)=4\sin^{-1}x-\frac{\pi}{2}$

Since $\sin^{-1}x$ is not a periodic function, $4\sin^{-1}x