If $$x = \sin^{-1}(\sin 10)$$ and $$y = \cos^{-1}(\cos 10)$$, then $$y - x$$ is equal to:

We recall the principal value ranges of the inverse trigonometric functions:

$$\sin^{-1}(t)\in\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right] \qquad\text{and}\qquad \cos^{-1}(t)\in\left[0,\,\pi\right].$$

Let $$\theta=10\;{\rm radians}.$$ Because $$2\pi\approx6.283$$ and $$4\pi\approx12.566,$$ the value $$\theta=10$$ satisfies

$$3\pi\;(\approx9.425)<10<4\pi\;(\approx12.566).$$

Evaluating $$x=\sin^{-1}(\sin10).$$

The sine function has period $$2\pi,$$ so $$\sin\theta=\sin(\theta-2n\pi)$$ for any integer $$n.$$ Choose $$n=1:$$

$$\theta-2\pi=10-2\pi\approx3.717.$$

This result still lies outside the principal interval $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right],$$ so subtract one more period:

$$\theta-4\pi=10-4\pi\approx-2.566.$$

The number $$-2.566$$ lies between $$-\pi$$ and $$-\dfrac{\pi}{2},$$ not yet in the principal range. Using $$\sin(-\alpha)=-\sin\alpha$$ and the symmetry $$\sin(\pi-\alpha)=\sin\alpha,$$ the standard result for this situation is

$$\sin^{-1}(\sin\theta)=(2n+1)\pi-\theta$$ when $$\theta\in\left[2n\pi+\dfrac{\pi}{2},\,2n\pi+\dfrac{3\pi}{2}\right].$$

Because $$10\in\left[2\pi+\dfrac{\pi}{2},\,2\pi+\dfrac{3\pi}{2}\right]$$ with $$n=1,$$ we obtain

$$x=(2\!\cdot\!1+1)\pi-10=3\pi-10.$$

Numerically, $$3\pi-10\approx-0.575,$$ which indeed lies inside $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].$$

Evaluating $$y=\cos^{-1}(\cos10).$$

The cosine function also has period $$2\pi,$$ and the standard formula is

$$ \cos^{-1}(\cos\theta)= \begin{cases} \theta-2m\pi, & \theta\in[2m\pi,\,2m\pi+\pi],\\[6pt] 2(m+1)\pi-\theta, & \theta\in[2m\pi+\pi,\,2m\pi+2\pi], \end{cases} $$ where $$m$$ is an integer.

Because $$10\in[3\pi,\,4\pi]=[2\!\cdot\!1\pi+\pi,\,2\!\cdot\!1\pi+2\pi]$$ with $$m=1,$$ we are in the second case. Substituting $$m=1$$ gives

$$y=2(m+1)\pi-\theta=2\!\cdot\!2\pi-10=4\pi-10.$$

Numerically, $$4\pi-10\approx2.566,$$ which lies in the required interval $$[0,\pi].$$

Finding $$y-x.$$ We have

$$ y-x=(4\pi-10)-(3\pi-10)=4\pi-10-3\pi+10=\pi. $$

Thus $$y-x=\pi.$$ Hence, the correct answer is Option B.