A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{2\pi }{3}\]
D) None of these
Correct Answer: D
Solution :
\[\alpha ={{\tan }^{-1}}\left\{ \frac{2\sqrt{\frac{{{\sin }^{2}}\theta }{4}-{{\sin }^{2}}\theta ({{\cos }^{2}}\theta -1)}}{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta -1} \right\}\] \[={{\tan }^{-1}}\infty \Rightarrow \alpha =\frac{\pi }{2}\]
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