If sin left( sin^{-1} frac{1}{5} + cos^{-1} x | vedclass

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(D) Given the equation: $\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1$. Taking $\sin^{-1}$ on both sides,we get: $\sin^{-1} \frac{1}{5}

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$\frac{1}{5}$

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(D) Given the equation: $\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x ight) = 1$. Taking $\sin^{-1}$ on both sides,we get: $\sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1}(1)$. Since $\sin^{-1}(1) = \frac{\pi}{2}$,the equation becomes: $\sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2}$. We know the identity $\sin^{-1} 	heta + \cos^{-1} 	heta = \frac{\pi}{2}$. Comparing this with our equation,we have $\sin^{-1} \frac{1}{5} = \frac{\pi}{2} - \cos^{-1} x$. Since $\frac{\pi}{2} - \cos^{-1} x = \sin^{-1} x$,we get $\sin^{-1} \frac{1}{5} = \sin^{-1} x$. Therefore,$x = \frac{1}{5}$.

If $\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1$,then $x$ is equal to

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