If sin^{-1} x + cos^{-1} y = frac{2pi}{5},the | vedclass

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(B) Given that,$\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5} \quad (1)$We know the fundamental identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ a

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$\frac{3\pi}{5}$

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(B) Given that,$\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5} \quad (1)$We know the fundamental identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $\sin^{-1} y + \cos^{-1} y = \frac{\pi}{2}$.From these,we can write $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y$.Substituting these into equation $(1)$:$(\frac{\pi}{2} - \cos^{-1} x) + (\frac{\pi}{2} - \sin^{-1} y) = \frac{2\pi}{5}$$\pi - (\cos^{-1} x + \sin^{-1} y) = \frac{2\pi}{5}$$\cos^{-1} x + \sin^{-1} y = \pi - \frac{2\pi}{5}$$\cos^{-1} x + \sin^{-1} y = \frac{3\pi}{5}$

If $\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5}$,then $\cos^{-1} x + \sin^{-1} y$ is

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