A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{8}\]
Correct Answer: B
Solution :
\[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +\beta }{2} \right)\] Þ \[\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=2\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\] \[\Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\] Þ \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{1}{3}\].
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