sin^{-1}x + sin^{-1}frac{1}{x} + cos^{-1}x + | vedclass

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(A) We know that $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.Given expression: $\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x +

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) We know that $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.Given expression: $\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x}$.Rearranging the terms,we get:$= (\sin^{-1}x + \cos^{-1}x) + (\sin^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{x})$.Using the identity $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$,we have:$= \frac{\pi}{2} + \frac{\pi}{2} = \pi$.Thus,the correct option is $A$.

$\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x} = $

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