The greatest real number C such that the inequality $|\tan x-\tan y|\geq C|x-y|$ holds for all $x,y \in \left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ 
Find the greatest real number $C$, such that the inequality $$|\tan x-\tan y|\geq C|x-y|$$ holds for all $$x,y \in \left(\frac{-\pi}{2},\frac{\pi}{2}\right)$$

All I've done so far is rearranged the inequality $$\bigg|\frac{\tan x-\tan y}{x-y}\bigg| \geq C$$
So since I want this to work for any $x,y$ in the interval I want to minimize $$\bigg|\frac{\tan x-\tan y}{x-y}\bigg|$$ On the interval $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ and let that equal $C$. Now this is where I'm drawing a blank. 
 A: According to MVT (mean value theorem), which @JyrkiLahtonen has mentioned and explained in his/her comment, you can find that, assuming $-\pi/2<x<y<\pi/2$
$$\frac{\tan y-\tan x}{y-x}=\sec^2 \xi\ge1$$
for some $\xi\in(x,y)$, this means that the requested $C$ is at least $1$, namely $C\ge1$. 
Then, fix $x=0$, and we get $\frac{\tan y-0}{y-0}$, which, when $y\to0$, is the derivative of $\tan$ at $0$. Thus $\forall \epsilon>0$, you can always find a $\delta>0$ s.t. 
$$1\le\frac{\tan y-0}{y-0}<1+\epsilon,\,\,\forall y\in(-\delta,0)\cup(0,\delta)$$
which by definition means that
$$\inf_{y\in(-\delta,0)\cup(0,\delta)} \frac{\tan y-0}{y-0}=1$$
Therefore 
$${1}\le C\le\inf_{y\in(-\delta,0)\cup(0,\delta)} \frac{\tan y-0}{y-0}=1$$
A: If $y=0$  then $\dfrac {|\tan x|} {x}| \ge C$ since
$|\dfrac {\tan x} {x}| \ge 1$ with $1$ being an infimum  then $C \le 1$
Hence we are left to prove inequality for $c=1$
W.L.O.G let $x\ge y$  then since $\tan x -x$  is increasing we have
$\tan x-x \ge \tan y-y \Rightarrow \dfrac {\tan x-\tan y} {x-y} \ge 1$
