If cos left(frac{alpha-beta}{2}right)=2 cos l | vedclass

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(C) Given,$\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$.Applying the componendo and dividendo rule:$\frac{\cos

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$\frac{1}{3}$

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(C) Given,$\cos \left(\frac{\alpha-\beta}{2}ight)=2 \cos \left(\frac{\alpha+\beta}{2}ight)$.Applying the componendo and dividendo rule:$\frac{\cos \left(\frac{\alpha-\beta}{2}ight) + \cos \left(\frac{\alpha+\beta}{2}ight)}{\cos \left(\frac{\alpha-\beta}{2}ight) - \cos \left(\frac{\alpha+\beta}{2}ight)} = \frac{2+1}{2-1}$.Using the sum-to-product formulas $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$ and $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:$\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}} = \frac{3}{1}$.$\cot \frac{\alpha}{2} \cot \frac{\beta}{2} = 3$.Therefore,$	an \frac{\alpha}{2} 	an \frac{\beta}{2} = \frac{1}{3}$.

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

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