If sin ^{-1} x+cos ^{-1} y=frac{3 pi}{10},the | vedclass

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(B) We are given the equation: $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$.Using the identity $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$ and $\cos

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$\frac{7\pi}{10}$

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(B) We are given the equation: $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$.Using the identity $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$ and $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$,we substitute these into the equation:$(\frac{\pi}{2} - \cos ^{-1} x) + (\frac{\pi}{2} - \sin ^{-1} y) = \frac{3 \pi}{10}$.Simplifying the left side:$\pi - (\cos ^{-1} x + \sin ^{-1} y) = \frac{3 \pi}{10}$.Rearranging the terms to solve for $\cos ^{-1} x + \sin ^{-1} y$:$\cos ^{-1} x + \sin ^{-1} y = \pi - \frac{3 \pi}{10}$.$\cos ^{-1} x + \sin ^{-1} y = \frac{10\pi - 3\pi}{10} = \frac{7 \pi}{10}$.

If $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$,then the value of $\cos ^{-1} x+\sin ^{-1} y$ is

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