The value of sin^{-1}left(cos frac{53pi}{5}ri | vedclass

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(D) Given expression: $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ We can write $\frac{53\pi}{5}$ as $10\pi + \frac{3\pi}{5}$. So,$\cos\left(10\pi + \

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$-\frac{\pi}{10}$

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(D) Given expression: $\sin^{-1}\left(\cos \frac{53\pi}{5}ight)$ We can write $\frac{53\pi}{5}$ as $10\pi + \frac{3\pi}{5}$. So,$\cos\left(10\pi + \frac{3\pi}{5}ight) = \cos\left(\frac{3\pi}{5}ight)$. Now,$\sin^{-1}\left(\cos \frac{3\pi}{5}ight) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}ight)ight)$ is incorrect,let us use $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$. $\sin^{-1}\left(\cos \frac{3\pi}{5}ight) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \frac{3\pi}{5}ight)ight)$ $= \sin^{-1}\left(\sin\left(\frac{5\pi - 6\pi}{10}ight)ight)$ $= \sin^{-1}\left(\sin\left(-\frac{\pi}{10}ight)ight)$ $= -\frac{\pi}{10}$.

The value of $\sin^{-1}\left(\cos \frac{53\pi}{5}\right)$ is

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