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(B) Given $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$.Expanding using the sine addition and subtraction formulas:$3(\sin \alpha \cos \beta + \cos \a

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$-5$

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(B) Given $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$.Expanding using the sine addition and subtraction formulas:$3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$$3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta$Rearranging the terms:$3 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta = -2 \cos \alpha \sin \beta - 3 \cos \alpha \sin \beta$$\sin \alpha \cos \beta = -5 \cos \alpha \sin \beta$Dividing both sides by $\cos \alpha \cos \beta$ (since $\alpha, \beta \in (0, \pi/2)$,$\cos \alpha, \cos \beta 
eq 0$):$\frac{\sin \alpha}{\cos \alpha} = -5 \frac{\sin \beta}{\cos \beta}$$	an \alpha = -5 	an \beta$Comparing this with $	an \alpha = k 	an \beta$,we get $k = -5$.

For $\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$,let $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$ and a real number $k$ be such that $\tan \alpha=k \tan \beta$. Then the value of $k$ is equal to :

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