If x = {sin ^{ - 1}}(sin 10) and y = {cos ^{ | vedclass

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(A) We know that the principal value range for $\sin^{-1}(\sin 	heta)$ is $[-\pi/2, \pi/2]$.Since $3\pi \approx 9.42$ and $10$ lies in the interval $

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$\pi $

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(A) We know that the principal value range for $\sin^{-1}(\sin 	heta)$ is $[-\pi/2, \pi/2]$.Since $3\pi \approx 9.42$ and $10$ lies in the interval $(2\pi, 3\pi)$,we use the property $\sin^{-1}(\sin 	heta) = 	heta - 3\pi$ for $	heta \in [2\pi, 3\pi]$.Thus,$x = \sin^{-1}(\sin 10) = 10 - 3\pi$.However,looking at the graph provided,the value is $3\pi - 10$.Let's re-evaluate: $10$ radians is in the third quadrant ($3\pi Actually,$3\pi Since $3\pi - 10 \in [-\pi/2, \pi/2]$,$x = 3\pi - 10$.For $y = \cos^{-1}(\cos 10)$,the range is $[0, \pi]$.Since $3\pi Thus,$y = 4\pi - 10$.Therefore,$y - x = (4\pi - 10) - (3\pi - 10) = \pi$.

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

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