If x takes a negative permissible value,then | vedclass

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Question

(A) Let $\sin^{-1} x = y$. Then,$x = \sin y$. Since $-1 \leq x < 0$,we have $-\frac{\pi}{2} \leq \sin^{-1} x < 0$,which implies $-\frac{\pi}{2} \leq y

Vedclass · Accepted Answer

$-\cos^{-1} \sqrt{1-x^{2}}$

Vedclass · Answer

(A) Let $\sin^{-1} x = y$. Then,$x = \sin y$. Since $-1 \leq x We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$. Since $y$ is in the interval $[-\frac{\pi}{2}, 0)$,$-y$ is in the interval $(0, \frac{\pi}{2}]$. Using the property $\cos(-y) = \cos y = \sqrt{1 - x^2}$,we get $-y = \cos^{-1}(\sqrt{1 - x^2})$. Therefore,$y = -\cos^{-1}(\sqrt{1 - x^2})$. Thus,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

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

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