Q117 / 180 · KCET 2009 · Mathematics
If ' $x$ ' takes negative permissible value, then $\sin ^{-1} \mathrm{x}$ is equal to
- Correct answerA$-\cos ^{-1} \sqrt{1-\mathrm{x}^{2}}$
- B$\cos ^{-1} \sqrt{\mathrm{x}^{2}-1}$
- C$\pi-\cos ^{-1} \sqrt{1-x^{2}}$
- D$\cos ^{-1} \sqrt{1-\mathrm{x}^{2}}$
Solution
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 $\quad-\frac{\pi}{2} \leq \mathrm{y} < 0$
$\Rightarrow \quad \frac{\pi}{2} \geq-y>0$
Hence, the correct answer is option: 'A'.