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(B) Let $\sin^{-1} x = y$. Then $x = \sin y$. Since $x$ is non-positive and permissible,we have $-1 \le x \le 0$. This implies $-\frac{\pi}{2} \le \si

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $\sin^{-1} x = y$. Then $x = \sin y$. Since $x$ is non-positive and permissible,we have $-1 \le x \le 0$. This implies $-\frac{\pi}{2} \le \sin^{-1} x \le 0$,so $-\frac{\pi}{2} \le y \le 0$. We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ for $y \in [0, \pi]$. Since $y \in [-\frac{\pi}{2}, 0]$,we have $-y \in [0, \frac{\pi}{2}]$. Thus,$\cos(-y) = \sqrt{1 - x^2}$. Since $\cos(-y) = \cos y$,we have $\cos y = \sqrt{1 - x^2}$. Taking the inverse cosine on both sides for the range $[0, \frac{\pi}{2}]$,we get $-y = \cos^{-1} \sqrt{1 - x^2}$. Therefore,$y = -\cos^{-1} \sqrt{1 - x^2}$. Hence,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

If $x$ takes non-positive permissible value,then $\sin^{-1} x =$

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