If cos^{-1} x + cos^{-1} y = 2pi,then sin^{-1 | vedclass

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(B) We know that for any $x, y \in [-1, 1]$,the range of $\cos^{-1} x$ is $[0, \pi]$.Given $\cos^{-1} x + \cos^{-1} y = 2\pi$.Since the maximum value

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$-\pi$

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(B) We know that for any $x, y \in [-1, 1]$,the range of $\cos^{-1} x$ is $[0, \pi]$.Given $\cos^{-1} x + \cos^{-1} y = 2\pi$.Since the maximum value of $\cos^{-1} x$ is $\pi$,the only way their sum can be $2\pi$ is if $\cos^{-1} x = \pi$ and $\cos^{-1} y = \pi$.This implies $x = \cos(\pi) = -1$ and $y = \cos(\pi) = -1$.Now,we need to find $\sin^{-1} x + \sin^{-1} y$.Substituting the values of $x$ and $y$:$\sin^{-1}(-1) + \sin^{-1}(-1) = -\frac{\pi}{2} - \frac{\pi}{2} = -\pi$.Alternatively,using the identity $\cos^{-1} t = \frac{\pi}{2} - \sin^{-1} t$:$(\frac{\pi}{2} - \sin^{-1} x) + (\frac{\pi}{2} - \sin^{-1} y) = 2\pi$.$\pi - (\sin^{-1} x + \sin^{-1} y) = 2\pi$.$\sin^{-1} x + \sin^{-1} y = \pi - 2\pi = -\pi$.

If $\cos^{-1} x + \cos^{-1} y = 2\pi$,then $\sin^{-1} x + \sin^{-1} y$ is equal to

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