sin ^{-1}(cos(sin ^{-1} x)) + cos ^{-1}(sin(c | vedclass

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(C) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Let $\sin^{-1} x = 	heta$,then $x = \sin 	heta$. Thus,$\cos(\sin^{-

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$\frac{\pi}{2}$

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(C) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Let $\sin^{-1} x = 	heta$,then $x = \sin 	heta$. Thus,$\cos(\sin^{-1} x) = \cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - x^2}$.So,$\sin^{-1}(\cos(\sin^{-1} x)) = \sin^{-1}(\sqrt{1 - x^2})$.Similarly,let $\cos^{-1} x = \phi$,then $x = \cos \phi$. Thus,$\sin(\cos^{-1} x) = \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - x^2}$.So,$\cos^{-1}(\sin(\cos^{-1} x)) = \cos^{-1}(\sqrt{1 - x^2})$.Now,the expression becomes $\sin^{-1}(\sqrt{1 - x^2}) + \cos^{-1}(\sqrt{1 - x^2})$.Using the identity $\sin^{-1} A + \cos^{-1} A = \frac{\pi}{2}$ where $A = \sqrt{1 - x^2}$,we get the result $\frac{\pi}{2}$.

$\sin ^{-1}(\cos(\sin ^{-1} x)) + \cos ^{-1}(\sin(\cos^{-1} x)) = \text{ . . . . . . }$

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