. Given that \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +B}{2} \right)\], then \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\]is equal to [AMU 2001]

. Given that \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \lef

1.	. Given that \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +B}{2} \right)\], then \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\]is equal to [AMU 2001]
A.	\[\frac{1}{2}\]
B.	\[\frac{1}{3}\]
C.	\[\frac{1}{4}\]
D.	\[\frac{1}{8}\]
Answer» C. \[\frac{1}{4}\]

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. Given that \[\cos \left( \frac{\alpha -\beta }{2} \right)=2\cos \left( \frac{\alpha +B}{2} \right)\], then \[\tan \frac{\alpha }{2}\tan \frac{\beta }{2}\]is equal to [AMU 2001]

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