For $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, let $3 \sin (\alpha+\beta)=2 \sin (\alpha-\bet

Question

For $\alpha, \beta \in\left(0, \frac{\pi}{2}ight)$, 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\

Professor PC Thomas Classes · Answer

(d): We have,\ $3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)$\ $\Rightarrow \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{2}{3}$\ $\Rightarrow \frac{\sin (\alpha+\beta)+\sin (\alpha-\beta)}{\sin (\alpha+\beta)-\sin (\alpha-\beta)}=\frac{5}{-1}$\ (Using componendo and dividendo)\ $\Rightarrow \frac{2 \sin \alpha \cos \beta}{2 \cos \alpha \sin \beta}=-5 \Rightarrow \frac{	an \alpha}{	an \beta}=-5$  \  $ \Rightarrow  	an \alpha = - tan \beta 	herefore k=5$