Prove that $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

(N/A) We have the left-hand side ($L$.$H$.$S$.) as: $\frac{\sin (x+y)}{\sin (x-y)}$.
Using the expansion formulas $\sin (x+y) = \sin x \cos y + \cos x \sin y$ and $\sin (x-y) = \sin x \cos y - \cos x \sin y$,we get:
$\frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$.
Dividing both the numerator and the denominator by $\cos x \cos y$,we get:
$\frac{\frac{\sin x \cos y}{\cos x \cos y} + \frac{\cos x \sin y}{\cos x \cos y}}{\frac{\sin x \cos y}{\cos x \cos y} - \frac{\cos x \sin y}{\cos x \cos y}} = \frac{\tan x + \tan y}{\tan x - \tan y}$.
Thus,$L$.$H$.$S$. = $R$.$H$.$S$.