Since $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
Using componendo and dividendo, we have
$$
\begin{aligned}
&\Rightarrow \frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{a+b+a-b}{a+b-a+b} \\
&\Rightarrow \frac{2 \sin \left(\frac{x+y+x-y}{2}\right) \cdot \cos \left(\frac{x+y-x+y}{2}\right)}{2 \cos \left(\frac{x+y+x-y}{2}\right) \cdot \sin \left(\frac{x+y-x+y}{2}\right)} \\
&\left.=\frac{2 a}{2 b}\left[\therefore \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} \text { and }\right] \sin x-\sin y=2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}\right] \\
&\Rightarrow \frac{\sin x \cdot \cos y}{\cos x \cdot \sin y}=\frac{a}{b} \Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}
\end{aligned}
$$