MCQ
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
- A$\frac{1}{2}$
- ✓$1\over3$
- C$\frac{1}{4}$
- D$\frac{1}{8}$
==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$
$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$
==> $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{1}{3}$.
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