To find the value of $k$, the given conditions are: $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$ And $\tan \alpha = k \tan \beta$ For the first equation, using the sum and difference formulas for sine, we can rewrite the equation as: $3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$ Simplifying this, we get: $3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta$ Rearranging the terms, we obtain: $5 \sin \beta \cos \alpha = - \sin \alpha \cos \beta$ Dividing both sides by $\sin \alpha \cos \beta$, we get: $\frac{5 \sin \beta \cos \alpha}{\sin \alpha \cos \beta} = -1$ Which simplifies to: $5 \tan \beta = - \tan \alpha$ So, taking the reciprocal, we have: $\tan \alpha = -5 \tan \beta$ Therefore, by comparing this equation with the given $\tan \alpha = k \tan \beta$, we find that $k = -5$. Thus, the value of $k$ is $-5$ .