If θ∈π2,3π2, then sin−1⁡(sin⁡θ) equals

If $\theta \in \left[\frac{\pi }{2},\frac{3\pi }{2}\right],$ then ${\mathrm{sin}}^{-1}\left(\mathrm{sin}\theta \right)$ equals

1. A

$\theta$

2. B

$\pi -\theta$

3. C

$2\pi -\theta$

4. D

$-\pi +\theta$

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Solution:

We have,

$\theta \in \left[\frac{\pi }{2},\frac{3\pi }{2}\right]⇒-\theta \in \left[-\frac{3\pi }{2},-\frac{\pi }{2}\right]⇒\pi -\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Also, $\mathrm{sin}\left(\pi -\theta \right)=\mathrm{sin}\theta$

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