Proving $x\leq \tan(x)$ How can i prove that $x\leq \tan(x)$ for any $x$ in $[0,\frac{\pi}{2})?$
 A: Define
$$f(x):=x-\tan x\implies f'(x)=1-\frac1{\cos^2x}=\frac{\cos^2x-1}{\cos^2x}\le0\implies f(x)$$
monotonic decreasing, and since $\;f(0)=0\;$, we get for $\;x\in[0,\pi/2)\;$ that
$$x-\tan x=f(x)\le f(0)=0\implies x\le \tan x$$
A: Proving this using calculus is likely a bit circular in logic. That's because $x < \tan x$ for $0<x<\pi/2$ is usually proved first, using geometry, before any derivatives of trig functions have been established. The inequality $x < \tan x$  is then used to obtain $(\sin x)/ x \to 1,$ which leads to the derivatives of all the trig functions.
The geometry argument goes like this: The sector of the unit circle determined by the arc $x$ has area $x/2.$ That sector has area smaller than the triangle with vertices $(0,0), (1,0), (1,\tan x);$ the picture makes that clear. The area of that triangle is $(1/2)\cdot 1 \cdot \tan x.$ Thus $x/2 < (\tan x) /2,$ which gives $x < \tan x$ as desired.
A: Hint. Let $x \in [0,\frac\pi2)$. Then, one has
$$
(x)'=1\leq 1+\tan^2 x=(\tan x)'
$$ and one may conclude with $0\leq \tan 0$.
A: Using Taylor's theorem for $x\in[0,\pi/2)$ write $\tan(x)=x+(sec\zeta\tan\zeta)\frac{x^2}{2}$ where $\zeta$ is some  number between $0$ and $x$. Since $(sec\zeta\tan\zeta)\geq 0$ we have that $\tan(x)\geq x.$
A: You could show, equivalently, that $\arctan(y)\le y$ for $y\ge 0$. For this the familiar fact
$$
\arctan(y) = \int_0^y {1\over 1+x^2}\,dx,\qquad y\ge 0
$$
will be useful.
A: You can't prove it because it is false:$$
\frac{3\pi}{4} > \tan\left( \frac{3\pi}{4} \right)=-1
$$
And in case you want the $\leq$ sign as in the title,
$$
\frac{\pi}{4} < \tan\left( \frac{3\pi}{4} \right)=1
$$
