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 . 

Consider fractions that are equivalent to $2/3$.
$$\frac{2}{3}\cdot\frac{2}{2} = \frac{4}{6}$$
$$\frac{2}{3}\cdot\frac{3}{3} = \frac{6}{9}$$
$$etc.$$
Conclusion ... you're allowed  to multiply by one.
Similarly, with the expression $\displaystyle\frac{\sin y\cos x+\cos y \sin x}{\sin y \cos x−\cos y \sin x}$, you're allowed to multiply by $\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}}$ because $$\displaystyle\frac{\frac{1}{\cos y \cos x}}{\frac{1}{\cos y \cos x}} = 1$$
And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows
$$\displaystyle\frac{\frac{\sin y\cos x+\cos y \sin x}{1}}{\frac{\sin y \cos x−\cos y \sin x}{1}}$$ before multiplying.
