Given, $\frac{\sin \left(x+y\right)}{\sin \left(x-y\right)}=\frac{a+b}{a-b}$

$\frac{\sin x.\cos y+\cos x.\sin y}{\sin x.\cos y-\cos x.\sin y}=\frac{a+b}{a-b}$

Use, componendo & dividendo

$\frac{\sin x.\cos y+\cos x.\sin y+\sin x.\cos y-\cos x.\sin y}{\sin x.\cos y+\cos x.\sin y-\sin x.\cos y+\cos x.\sin y}=\frac{a+b+a-b}{a+b-a+b}$

$\frac{2\sin x.\cos y}{2\cos x.\sin y}=\frac{2a}{2b}$

$\frac{{\displaystyle \frac{\sin x}{\cos x}}}{{\displaystyle \frac{\sin y}{\cos y}}}=\frac{a}{b}$

$\frac{\tan x}{\tan y}=\frac{a}{b}$