If $\tan A=\frac{x \sin B}{1-x \cos B}$ and $\tan B=\frac{y \sin A}{1-y \cos A}$ then $\frac{\sin A}{\sin B}=\_\_\_\_$

$\tan A=\frac{\sin A}{\cos A}=\frac{xsinB}{1-x\cos B}$
$\sin A-x\sin A\cos B=x\cos A\sin B$
$\sin A=x\left(\sin A\cos B+\cos A\sin B\right)$
$\sin A=x\sin \left(A+B\right)$ Similarly 
$\sin B=y\sin \left(A+B\right)$ 
$\frac{\sin A}{\sin B}=\frac{x\sin \left(A+B\right)}{y\sin \left(A+B\right)}=\frac{x}{y}$