If 3 sin alpha = 5 sin beta,then tan left(fra | vedclass

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(D) Given that $3 \sin \alpha = 5 \sin \beta$,we can write $\frac{\sin \alpha}{\sin \beta} = \frac{5}{3}$.Applying Componendo and Dividendo,we get:$\f

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$4$

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(D) Given that $3 \sin \alpha = 5 \sin \beta$,we can write $\frac{\sin \alpha}{\sin \beta} = \frac{5}{3}$.Applying Componendo and Dividendo,we get:$\frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.Using the sum-to-product formulas:$\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2}ight) \cos \left(\frac{\alpha - \beta}{2}ight)$$\sin \alpha - \sin \beta = 2 \cos \left(\frac{\alpha + \beta}{2}ight) \sin \left(\frac{\alpha - \beta}{2}ight)$Substituting these into the ratio:$\frac{2 \sin \left(\frac{\alpha + \beta}{2}ight) \cos \left(\frac{\alpha - \beta}{2}ight)}{2 \cos \left(\frac{\alpha + \beta}{2}ight) \sin \left(\frac{\alpha - \beta}{2}ight)} = 4$$	an \left(\frac{\alpha + \beta}{2}ight) \cot \left(\frac{\alpha - \beta}{2}ight) = 4$Since $\cot 	heta = \frac{1}{	an 	heta}$,we have $	an \left(\frac{\alpha + \beta}{2}ight) \div 	an \left(\frac{\alpha - \beta}{2}ight) = 4$.

If $3 \sin \alpha = 5 \sin \beta$,then $\tan \left(\frac{\alpha + \beta}{2}\right) \div \tan \left(\frac{\alpha - \beta}{2}\right) = $

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