જો cos left( {frac{{alpha - beta }}{2}} right | vedclass

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Question

(b) $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} \right)$

==> $\cos \frac{\alpha }{2}\cos \fr

Vedclass · Accepted Answer

$1\over3$

Vedclass · Answer

(b) $\cos \left( {\frac{{\alpha - \beta }}{2}} ight) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} ight)$

==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$

$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$

==> $	an \frac{\alpha }{2}	an \frac{\beta }{2} = \frac{1}{3}$.

જો $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, તો $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = . . .$

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