Proof that $\tan(x)\leq{}x-\frac{\pi}{4}+\tan(\frac{\pi}{4})$ using the Mean Value Theorem

Question

We're asked to proof the above inequality $\forall{}x\in(\frac{-\pi}{2},\frac{\pi}{4}]$. Although am a bit confused with the fact that $\tan(\frac{-\pi}{2})$ is undefined. And thus am stuck when applying the mean value theorem to proof the statement. What interval should I be taking in order to apply the theorem? Thanks in advance.

My attempt:

Let $f(t)=\tan(t)$ such that $t\in(\frac{-\pi}{2},x]$ whereas $x\leq{}\frac{\pi}{4}$ then, because $f$ is continuous on the closed interval and differentiable on the open interval, $\exists{}c\in(-\frac{\pi}{2},x)$ such that the difference of the quotient is equal to $f'(c)$ but then again, I can't take $f(-\frac{\pi}{2})$

Answer 1

If $x=\frac\pi4$, then $\tan(x)=x-\frac\pi4+\tan\left(\frac\pi4\right)$. On the other and, if $x\in\left(-\frac\pi2,\frac\pi4\right)$, then$$\frac{\tan(x)-x-\left(\tan\left(\frac\pi4\right)-\frac\pi4\right)}{x-\frac\pi4}=\tan'(c)-1,$$for some $c$ between $x$ and $\frac\pi4$. But $\tan'(c)-1+=\tan^2(c)>0$. Since $x-\frac\pi4<0$, you deduce that $\tan(x)-x-\left(\tan\left(\frac\pi4\right)-\frac\pi4\right)<0$. In other words, $ \tan(x)<x-\frac\pi4+\tan\left(\frac\pi4\right)$.

Answer 2

This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $\tan x$ instead of $\tan x-x$.

Rewrite the inequality to be proved first as

$${\pi\over4}-x\le\tan\left(\pi\over4\right)-\tan x$$

and then, since we're only concerned with $x\lt{\pi\over4}$ (after noting that the inequality is a trivial equality for $x={\pi\over4}$), as

$$1\le{\tan\left(\pi\over4\right)-\tan x\over{\pi\over4}-x}$$

Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $c\in(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-\pi/2\lt x$ (and, here, less than $\pi/4$), so we have

$${\tan\left(\pi\over4\right)-\tan x\over{\pi\over4}-x}=\sec^2c$$

for some $c\in(x,\pi/4)$. Since $\sec^2c\ge1$ for all $c$ (in the appropriate domain), we're done.