Trigonometric Identities Question 319
Question: . Given that $ \cos ( \frac{\alpha -\beta }{2} )=2\cos ( \frac{\alpha +B}{2} ) $ , then $ \tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to
[AMU 2001]
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{3} $
C) $ \frac{1}{4} $
D) $ \frac{1}{8} $
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Answer:
Correct Answer: B
Solution:
$ \cos ( \frac{\alpha -\beta }{2} )=2\cos ( \frac{\alpha +\beta }{2} ) $
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} $ .
BETA
