If $x={\sin }^{-1}\left(\sin 10\right) \mathrm{and} y={\cos }^{-1}\left(\cos 10\right)$ then y - x =

$\begin{array}{l}3\pi <10<4\pi \Rightarrow 10\text{ lies in }{Q}_4\\ \therefore x={\sin }^{-1}(\sin 10)\\ x={\sin }^{-1}(\sin (3\pi -10\left)\right)\\ =3\pi -10\\ \therefore y={\cos }^{-1}(\cos 10)={\cos }^{-1}(\cos (4\pi -10\left)\right)=4\pi -10\\ \therefore y-x=(4\pi -10)-(3\pi -10)=\pi \\ y-x=\pi \end{array}$