Given that cos left( frac{alpha - beta}{2} ri | vedclass

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(B) Given the equation: $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$.Using the expansion formulas $\

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$\frac{1}{3}$

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(B) Given the equation: $\cos \left( \frac{\alpha - \beta}{2} ight) = 2\cos \left( \frac{\alpha + \beta}{2} ight)$.Using the expansion formulas $\cos(A - B) = \cos A \cos B + \sin A \sin B$ and $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \left( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2} ight)$.Expanding the right side:$\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$.Rearranging the terms to group $\sin$ and $\cos$ products:$3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$.Dividing both sides by $3 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$,we obtain:$\frac{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{1}{3}$.Therefore,$	an \frac{\alpha}{2} 	an \frac{\beta}{2} = \frac{1}{3}$.

Given that $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ is equal to

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