For \alpha, \beta \in(0, \pi / 2), let 3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta) and a real number k be such that an \alpha=k an \beta. Then, the value of k is equal to

Question

For \alpha, \beta \in(0, \pi / 2), let 3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta) and a real number k be such that 	an \alpha=k 	an \beta. Then, the value of k is equal to

Accepted Answer

Answer: C. -5 To find the value of k, the given conditions are: 3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta) And 	an \alpha = k 	an \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 	an \beta = - 	an \alpha So, taking the reciprocal, we have: 	an \alpha = -5 	an \beta Therefore, by comparing this equation with the given 	an \alpha = k 	an \beta, we find that k = -5. Thus, the value of k is -5.

For $$\alpha, \beta \in(0, \pi / 2)$$, 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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