A) x
B) \[-x\]
C) \[\pi +x\]
D) \[\pi -x\]
Correct Answer: D
Solution :
We have \[\frac{\pi }{2}\le x\le \frac{3\pi }{2}\] \[\Rightarrow \,\,\frac{-\pi }{2}\le x-\pi \le \frac{\pi }{2}\,\,\Rightarrow \,\,\frac{-\pi }{2}\le \pi -x\le \frac{\pi }{2}\] \[\Rightarrow \,\,{{\sin }^{-1}}\{\sin \,(\pi -x)\}=\pi -x\].
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