A) \[\pi /4\]
B) \[\pi /2\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: A
Solution :
\[2\sin 2\alpha =\tan \beta +\cot \beta \] \[2.\frac{1}{\cos 2\alpha }=\frac{\sin \beta }{\cos \beta }+\frac{\cos \beta }{\sin \beta }\] \[\frac{2}{\cos 2\alpha }=\frac{{{\sin }^{2}}\beta +{{\cos }^{2}}\beta }{\cos \beta \sin \beta }\] \[\cos 2\alpha =\sin 2\beta \] \[\cos 2\alpha =\cos \left( \frac{\pi }{2}-2\beta \right)\] \[2\alpha =\frac{\pi }{2}-2\beta \Rightarrow \alpha +\beta =\pi /4\]
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