Trig inequality $|\tan x-\tan y|\leq \frac{1}{\cos^2 c}|x-y|$

Question

Let $0<c<\frac{\pi}{2}$. Show that the following holds for all $x,y\in [-c,c]$:

$$|\tan x-\tan y|\leq \frac{1}{\cos^2 c}|x-y|$$

I've thought about applying this lemma but I haven't been able to figure out so far how to exactly do that in the above case.

Answer 1

This follows from the mean value theorem and the fact that $\tan'=\frac1{\cos^2}$.

Answer 2

We know that $\cos$ decreases on $\left(0,\frac{\pi}{2}\right)$ and by the Lagrange's theorem there is $\theta\in(0,1),$ for which: $$|\tan{x}-\tan{y}|=\frac{|x-y|}{\cos^2\theta c}\leq\frac{|x-y|}{\cos^2 c}$$