Limit of $\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $ without l'hopital's Find limit of 

$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} $$

I started by defining $\ x = y + \frac{\pi}{4} $ 
$$\ \lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4} = \lim_{y \to 0} \frac{\tan (y+ \pi/4)-1}{y}= ? $$
This is exactly the same question where the solution suggested also used $\ x = y + \pi/4 $ yet I couldn't understand this equality:  $$\ \tan(x) = \tan(y + \pi/4) =  \frac{\tan y -1 }{1- \tan y} $$ and then how is that exactly equal to limit of $$\ \lim_{y \to 0}\frac{1}{y} \left( 1-\frac{\tan y +1}{1 - \tan y}\right)$$

$$\frac{\tan\left(y+\dfrac\pi4\right)-1}y=\frac{\dfrac{\tan y+\tan\dfrac \pi4}{1-\tan y\tan \dfrac \pi4}-1}y=2\frac{\tan y}{y}\frac1{1-\tan y}\to 2$$
