<s>[INST] Proving trigonometric identity: $ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $ 
Prove that 
  $$ \frac{\sin (y+x)}{\sin (y-x)} = \frac{\tan y + \tan x}{\tan y - \tan x} $$ 

This is my working - 
$$\text{LHS} = \frac{\sin (y+x)}{\sin (y-x)} = \frac{\sin y \cos x + \cos y \sin x}{\sin y \cos x - \cos y \sin x} $$
I used the compound angle relations formula to do this step. 
However , now I'm stuck. 
I checked out the answer and the answer basically carried on my step by dividing it by
$ \frac{\cos x \cos y}{\cos x \cos y} $ 
If I'm not wrong , we cannot add in our own expression right? Because the question is asking to prove left is equal to right. Adding in our own expression to the left hand side will be not answering the question, I feel . 
 [/INST] $$\frac{\tan x+\tan y}{\tan x-\tan y} =\frac{\frac{\sin y\cos x+\cos y\sin x}{\cos x\cos y}}{\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}}= \frac{\sin y\cos x+\cos y\sin x}{\sin y\cos x-\cos y\sin x}=\frac{\sin (y+x)}{\sin (y-x)} $$
which gives the results since we know that 
$$ \sin(x-y) =\sin y\cos x-\cos y\sin x $$and 
$$ \sin(x+y) =\sin y\cos x+\cos y\sin x $$
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