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 $$\ \lim_{y \to 0}\frac{1}{y} \left( 1-\frac{\tan y +1}{1 - \tan y}\right)$$

As an alternative without derivatives, let $x=\frac{\pi}{4}-y$ then
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan x-1}{x-\pi/4}=\lim_{y \to 0} \frac{\tan \left(\frac{\pi}{4}-y\right)-1}{-y}=\lim_{y \to 0} \frac{1-\frac{\cos \left(2y\right)}{1+\sin \left(2y\right)}}{y}=\lim_{y \to 0} \frac{1-\cos \left(2y\right)+\sin \left(2y\right)}{y(1+\sin \left(2y\right))}=$$
$$=\lim_{y \to 0} \frac{1}{1+\sin \left(2y\right)}\left(4y\frac{1-\cos \left(2y\right)}{4y^2}+2\frac{\sin \left(2y\right)}{2y}\right)=2$$
