Prove that $\tan(x) \gt x$ for any $x$, $x\in(0,π/2)$

Question

I'mm trying to prove that with Lagrange sentence,

$F(x)= \tan(x)$

$G(x)= x$

I was trying to show that for any $x\in(0,π/2)$, $F'(x)=G'(X)$.

Then according to Lagrange: $G(X) = C\cdot F(X)$ which means that $F(X)\lt G(X)$ for any $x\in(0,π/2)$.

Answer 1

Hint:

Do you know the Maclaurin series for tan(x)

and where it converges?

Answer 2

You can simply use the fact that, for $x=0, x=tan x$ and$ (tan x)' \ge x' $for all $\pi/2 \ge x\ge0$. Consider $f=tan x-x. f'=sec^2x-1 \ge 0$ for all $\pi/2 \ge x\ge0$ and f(0)=0. So f is increasing, therefore $tan x-x \ge 0.$

Answer 3

By integration from $0$ to $x$,

$$\sec^20=1,\sec^2x>1\implies \tan x> x.$$

Answer 4

Observation: We are working in the interval $\big(0, \frac{\pi}{2}\big)$.

Let $f(x):=\tan(x)$ and $g(x):=x$

Observe that $$\begin{align*}f'(x)&=\frac{1}{\cos^2(x)}\tag{1}\\&>1=g'(x)\end{align*}$$

Notice, furthermore, that $$f(0)=g(0)$$

Can you finish now?

---

Addendum

There is also a nice geometric proof

It follows from the areal comparison of $$\triangle ABC=\frac{\tan(\alpha)}{2}>\text{circular sector }ABD=\frac{\alpha}{2\pi}\cdot(\pi\cdot1^2)$$ Or equivalently $$\frac{\tan(\alpha)}2>\frac\alpha2\iff \tan(\alpha)>\alpha$$