proving an inequality by the mean value theorem Use the Mean Value Theorem to prove that for all $a,b\in(\frac\pi8,\frac\pi4)$ we have $$|\tan(2a)-\tan(2b)|\geq4|a-b|$$ If using a theorem, justify each of the hypothesis of the theorem before applying it.
This is what I think it is but I am really unsure and would like to see a detailed solution please.
Define function $f:[a,b]\rightarrow\mathbb{R}$ by $f(x)=\tan(2x)$ with $a,b\in(\frac\pi8,\frac\pi4)$. 
Note that since, $\tan(x)$ and $2x$ are continuous functions on the real line, then the composed function $\tan(2x)$ is continuous by the algebra of continuous functions. Hence $f$ is continuous on interval $[a,b]$. Also since $\tan(x)$ and $2x$ are differentiable functions, by the algebra of differentiable functions, the composed function $\tan(2x)$ is differentiable. Hence $f$ is differentiable on $(a,b)$.
Applying the MVT, there exists $x\in(a,b)$ such that $$\frac{f(b)-f(a)}{b-a}=\frac{\tan(2b)-\tan(2a)}{b-a}=\frac2{\cos^2(2c)}=f'(c)$$
Now I would just say $0\leq\cos^2(2c)\leq1$ but looking at what $a$ and $b$ can be, this is $0<\cos^2(2c)<1$ but the question has a more or equal to. I don't understand how you can get that.
 A: If $c\in(\frac\pi8,\frac\pi4)$, then $2c\in(\frac\pi4,\frac\pi2)$ and:
$$ 0 =\cos^2\left(\frac{\pi}{2}\right) <\cos^2(2c) < \cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2}. $$
Hence:
$$ \frac{2}{\cos^2(2c)} > 4.$$
Edit: 
If $ a= b$, then
$$ |\tan (2a) - \tan (2b)| = 0 = 4|a-b|. $$
If $a\not= b$, proceed as you did with the application of MVT to obtain:
$$ |\tan (2a) - \tan (2b)| > 4|a-b|. $$
So, no matter what the relationship between $a$ and $b$ is, as long as they are both in $(\frac\pi8,\frac\pi4)$ we have
$$ |\tan (2a) - \tan (2b)| \geq 4|a-b|. $$
Edit2:
If $a> b$, we show 
$$ \frac{\tan (2a) - \tan (2b)}{a-b} =  \frac{2}{\cos^2(2c)}, $$
for some $c\in (b,a)$.
If $a<b$, we show 
$$ \frac{\tan (2b) - \tan (2a)}{b-a} =  \frac{2}{\cos^2(2c)}, $$
for some $c\in (a,b)$.
Concisely, using modulus, we say that for $a\not=b$ we have:
$$ \frac{|\tan (2b) - \tan (2a)|}{|b-a|} =  \frac{2}{\cos^2(2c)}, $$
for some $c$ in $(a,b)$ or $(b,a)$.
Function $\tan$ is strictly increasing on $(\frac\pi4,\frac\pi2)$. 
A: Since $a,b\in(\frac\pi8,\frac\pi4)$, we have $2c\in(\frac\pi4,\frac\pi2)$. Then
$$
\cos\frac\pi2<\cos 2c<\cos\frac\pi4\implies\cos^22c\le\frac{1}{2}
$$
and
$$
\frac{1}{\cos^22c}\ge2.
$$
