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

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

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$\frac{1}{3}$

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(B) Given: $\cos \left( \frac{\alpha - \beta}{2} ight) = 2\cos \left( \frac{\alpha + \beta}{2} ight)$ Using the expansion formulas $\cos(A-B) = \cos A \cos B + \sin A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \left( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - \sin \frac{\alpha}{2} \sin \frac{\beta}{2} ight)$ $\cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = 2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} - 2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}$ Rearranging the terms: $3 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} = \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$ Dividing both sides by $3 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$: $\frac{\sin \frac{\alpha}{2} \sin \frac{\beta}{2}}{\cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \frac{1}{3}$ $	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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