How to prove $|\sin x_1-\sin x_2|\leqslant |\tan x_1-\tan x_2|$ is correct?

Question

I really have no clue how to solve this problem, can anyone help please?

Prove the equation:

$$|\sin x_1-\sin x_2| \leqslant|\tan x_1-\tan x_2|$$

I tried to use this rule: If a function is continuous in $[a,b]$, and has derivative in $(a,b)$ then, there is a point like $c$ that : $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ But I failed...

Answer 1

In light of the mean value theorem and that fact that the derivatives of $\sin$ and $\tan$ are bounded above and below, respectively, by $1$ you have $$|\sin x_1 - \sin x_2| \le |x_1 - x_2| \le |\tan x_1 - \tan x_2|$$ provided you are working on an interval where the mean value theorem is valid.