Proving that $\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\tan y$ Edit: got it, silly mistakes :)
I need to prove that $\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\tan y$
$$=\frac{\tan x+\tan y-\tan x+\tan^2x\tan y}{1-\tan x\tan y+\tan^2x+\tan x\tan y}$$
$$=\frac{\tan y+\tan^2x\tan y}{1+\tan^2x}$$
$$=\frac{(\tan y)(1+\tan^2x)}{(1+\tan^2x)}$$
$$\boxed{=\tan y}$$
Any hints/help appreciated.
Thanks

Hint: Recall that 
$$\tan(u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v}.\tag{1}$$ Let $u=x+y$ and $v=x$. 
Remark: If Identity (1) requires proof, use
$$\tan(u-v)=\frac{\sin(u-v)}{\cos(u-v)}=\frac{\sin u\cos v-\cos u\sin v}{\cos u\cos v+\sin u\sin v},$$
and divide top and bottom by $\cos u\cos v$. 
