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If ' $x$ ' takes negative permissible value, then $\sin ^{-1} \mathrm{x}$ is equal to
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Verified Answer
The correct answer is:
$-\cos ^{-1} \sqrt{1-\mathrm{x}^{2}}$
Let $\sin ^{-1} x=y$. Then, $x=\sin y$
Since, $-1 \leq x < 0$, therefore $-\frac{\pi}{2} \leq \sin ^{-1} x < 0$
$\Rightarrow \quad-\frac{\pi}{2} \leq \mathrm{y} < 0$
Now, $\cos y=\sqrt{1-\sin ^{2} y}$
$\Rightarrow \quad \cos y=\sqrt{1-x^{2}}$ for $0 \leq y \leq \pi$
But $\quad-\frac{\pi}{2} \leq \mathrm{y} < 0$
$\Rightarrow \quad \frac{\pi}{2} \geq-y>0$
Since, $-1 \leq x < 0$, therefore $-\frac{\pi}{2} \leq \sin ^{-1} x < 0$
$\Rightarrow \quad-\frac{\pi}{2} \leq \mathrm{y} < 0$
Now, $\cos y=\sqrt{1-\sin ^{2} y}$
$\Rightarrow \quad \cos y=\sqrt{1-x^{2}}$ for $0 \leq y \leq \pi$
But $\quad-\frac{\pi}{2} \leq \mathrm{y} < 0$
$\Rightarrow \quad \frac{\pi}{2} \geq-y>0$
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