If 5 sin α=3 sin

(c) $4 \tan \beta$

We have,

$5 \sin \alpha=3 \sin (\alpha+2 \beta)$

$\Rightarrow \frac{5}{3}=\frac{\sin (\alpha+2 \beta)}{\sin \alpha}$

$\Rightarrow \frac{5-3}{5+3}=\frac{\sin (\alpha+2 \beta)-\sin \alpha}{\sin (\alpha+2 \beta)+\sin \alpha} \quad$ (Using componendo and dividendo)

$\Rightarrow \frac{2}{8}=\frac{\sin (\alpha+2 \beta)-\sin \alpha}{\sin (\alpha+2 \beta)+\sin \alpha}$

$\Rightarrow \frac{1}{4}=\frac{2 \cos \frac{\alpha+2 \beta+\alpha}{2} \sin \frac{\alpha+2 \beta-\alpha}{2}}{2 \sin \frac{\alpha+2 \beta+\alpha}{2} \cos \frac{\alpha+2 \beta-\alpha}{2}}$

$\Rightarrow \frac{1}{4}=\frac{\cos (\alpha+\beta) \sin \beta}{\sin (\alpha+\beta) \cos \beta}$

$\Rightarrow \frac{1}{4}=\cot (\alpha+\beta) \tan \beta$

$\Rightarrow \frac{1}{4}=\frac{1}{\tan (\alpha+\beta)} \tan \beta$

$\therefore \tan (\alpha+\beta)=4 \tan \beta$