Solution:
As per the question,
$\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
$\text { Since, } \sin (A+B)=\sin A \cos B+\cos A \sin B$
$\therefore \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$
$\Rightarrow \frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}=\frac{a+b}{a-b}$
Using the componendo-dividendo rule,
We have,
$\Rightarrow \frac{(\sin x \cos y+\cos x \sin y)+(\sin x \cos y-\cos x \sin y)}{(\sin x \cos y+\cos x \sin y)-(\sin x \cos y-\cos x \sin y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}$
$\Rightarrow \frac{2 \sin x \cos y}{2 \cos x \sin y}=\frac{2 a}{2 b}$
$\Rightarrow\left(\frac{\sin x}{\cos x}\right)\left(\frac{\cos y}{\sin y}\right)=\frac{a}{b}$
$\text { As, } \tan A=(\sin A) / \cos (A)$
$\Rightarrow \tan x\left(\frac{1}{\tan y}\right)=\frac{a}{b}$
$\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b}$