Proving $x\leq \tan(x)$ How can i prove that $x\leq \tan(x)$ for any $x$ in $[0,\frac{\pi}{2})?$
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

$$x-\tan x=f(x)\le f(0)=0\implies x\le \tan x$$
You can't prove it because it is false:$$
\frac{3\pi}{4} &gt; \tan\left( \frac{3\pi}{4} \right)=-1
$$
And in case you want the $\leq$ sign as in the title,
$$
\frac{\pi}{4} &lt; \tan\left( \frac{3\pi}{4} \right)=1
$$