If sin^{-1} x + sin^{-1} y = frac{2pi}{3},the | vedclass

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(B) We know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Therefore,$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\s

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$\frac{\pi}{3}$

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(B) We know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Therefore,$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y$.Substituting these into the given equation:$(\frac{\pi}{2} - \cos^{-1} x) + (\frac{\pi}{2} - \cos^{-1} y) = \frac{2\pi}{3}$$\pi - (\cos^{-1} x + \cos^{-1} y) = \frac{2\pi}{3}$$\cos^{-1} x + \cos^{-1} y = \pi - \frac{2\pi}{3}$$\cos^{-1} x + \cos^{-1} y = \frac{\pi}{3}$.

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$,then $\cos^{-1} x + \cos^{-1} y = $

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