JEE Main 2025 Apr 4 Shift 1, Mathematics Q13: Inverse Trigonometric Functions

Let $\theta ={\sin }^{-1}x$, so $x=\sin \theta$ and $\sqrt{1-x^2}=\cos \theta$ (since the given range of $x$ corresponds to $-\frac{\pi }{6}<\theta <\frac{\pi }{4}$, where $\cos \theta >0$).

The argument of the outer ${\sin }^{-1}$ becomes $\frac{\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta =\sin \left(\theta +\frac{\pi }{6}\right)$.

For $-\frac{\pi }{6}<\theta <\frac{\pi }{4}$, $\theta +\frac{\pi }{6}$ lies in $\left(0,\frac{5\pi }{12}\right)\subset \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, so ${\sin }^{-1}\sin \left(\theta +\frac{\pi }{6}\right)=\theta +\frac{\pi }{6}$.

Therefore the expression equals ${\sin }^{-1}(x)+\frac{\pi }{6}$.