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

$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$
$$a=\tan x\;\;and\;\;b=\tan y$$
$$\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\frac{\frac{a+b}{1-ab}-a}{1+\frac{a+b}{1-ab}a}$$
$$=\frac{a+b-a(1-ab)}{1-ab+a(a+b)}$$
$$=\frac{b(a^2+1)}{a^2+1}$$
$$=b=\tan y$$
