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.

score 2 · Accepted Answer · answered Jan 13 '20 at 19:14

2

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

answered Jan 13 '20 at 19:14

José Carlos Santos

427,504

Thank you, so now I have that $|\tan x-\tan y|=|x-y||\tan' \xi|$ for some $\xi$ but how do I continue from here? – Jan 13 '20 at 19:22
$$\lvert\tan'\xi\rvert=\left\lvert\frac1{\cos^2\xi}\right\rvert=\frac1{\cos^2\xi}.$$ – José Carlos Santos Jan 13 '20 at 19:23
Yes, that's exactly my problem: How do I make the connection to the given $c$? – Jan 13 '20 at 19:24
Since $\frac1{\cos^2}$ is increasing on $\left[0,\frac\pi2\right]$, $$\frac1{\cos^2\xi}\leqslant\frac1{\cos^2c}.$$ – José Carlos Santos Jan 13 '20 at 19:29

score 0 · Answer 2 · answered Jan 13 '20 at 19:20

0

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: $$|\tan{x}-\tan{y}|=\frac{|x-y|}{\cos^2\theta c}\leq\frac{|x-y|}{\cos^2 c}$$

answered Jan 13 '20 at 19:20

Michael Rozenberg

194,933

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

2 Answers2