A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: D
Solution :
[d] \[\tan \,\,(\sin x)=\tan \left( \frac{\pi }{2}-\sin x \right)\] \[\therefore \,\,\sin x=\frac{\pi }{2}-\sin x+n\pi ,\,n\in I\] \[\Rightarrow \,\,\sin x=\frac{\pi }{4}+\frac{n\pi }{2},n\in I\] \[\Rightarrow \,\,\sin x=\frac{\pi }{4}\] or \[-\frac{\pi }{4}\] as \[\left| \frac{n\pi }{2}+\frac{\pi }{4} \right|>1\forall n\in I-\{-1,0\}\] Hence, there will be one value of x in each quadrant.
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