why does $\tan x > x$ mean $\tan x − x$ is increasing ? If the right side x is transferred to the left side, the inequality is equal to $\tan x-x>0$.
I think if the x is transferred to the left side, we need to prove that $\tan x-x>0$ for $0<x<\frac\pi2$, which means the function of $\tan x-x$ is above the $x$- axis. enter image description here
I'll approach with derivations.
Let $f(x)=\tan x-x$, then $f'(x)=\frac{1}{\cos^2x}-1=\tan^2x>0$ when $0<x<\frac\pi2$.
So $f(x)$ increases at that range.