Note $10$ radians. Since $3\pi\approx9.42$ and $4\pi\approx12.57$: $3\pi<10<4\pi$.

For $x$: $\sin^{-1}(\sin10)$. Principal range $[-\pi/2,\pi/2]$. $10\approx10-3\pi\approx0.58$, but $10-(3\pi)\approx0.58<\pi/2$? $3\pi\approx9.42$, $10-3\pi\approx0.58$. Since $\sin(10)=\sin(\pi-(10-3\pi))=\sin(4\pi-10)$... more carefully: $10 = 3\pi + (10-3\pi)$, $\sin(10)=\sin(3\pi+(10-3\pi))=-\sin(10-3\pi)$. So $x=\sin^{-1}(-\sin(10-3\pi))=-(10-3\pi)=3\pi-10$.

For $y$: $\cos^{-1}(\cos10)$, range $[0,\pi]$. $10-3\pi\approx0.58\in[0,\pi]$, so $y=10-3\pi$... wait: $\cos(10)=\cos(4\pi-10)$ and $4\pi-10\in[0,\pi]$. So $y=4\pi-10$.

$y-x=(4\pi-10)-(3\pi-10)=\mathbf{\pi}$