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3.Trigonometrical Ratios, Functions and Identities
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यदि $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} \right)$, तो $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}$ का मान होगा
A
$\frac{1}{2}$
B
$1\over3$
C
$\frac{1}{4}$
D
$\frac{1}{8}$
Solution
(b) $\cos \left( {\frac{{\alpha – \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} \right)$
==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} – 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$
$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$
==> $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{1}{3}$.
Standard 11
Mathematics
