The value of $sin^{-1} (sin 10)$, is

$3\pi - 10$

We know that $sin^{-1}(sin \theta) = \theta ,$ if $-\frac{\pi}{2} ≤ θ ≤ \frac{\pi}{2}$.

 Here, θ = 10 radians which does not lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

But, $3\pi - \theta$ i.e $3\pi - 10$ lies, netween $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

Also, $sin(3\pi - 10) = sin 10.$

∴ $sin^{-1}(sin 10)=sin^{-1}(sin (3\pi - 10))= 3\pi - 10$