If $x$ takes a negative permissible value, then $\sin^{-1}x$ is equal to

Question

$\cos^{-1}\left(\sqrt{1 - x^2}\right)$

$-\cos^{-1}\left(\sqrt{1 - x^2}\right)$

$\cos^{-1}\left(\sqrt{x^2 - 1}\right)$

$\pi - \cos^{-1}\left(\sqrt{1 - x^2}\right)$

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