MCQ
If $\sin ^{-1} x=y,$ then
- A$-\frac{\pi}{2} < y < \frac{\pi}{2}$
- B$0 \leq \mathrm{y} \leq \pi$
- ✓$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
- D$0 < y < \pi$
We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
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