Answer
Prove that $f(x)$ is increasing on $\left(0,\frac{\pi}{2}\right)$
Work Step by Step
$f(x)$ = $\tan x-x$
$f'(x)$ = $\sec^{2}x-1$ $\gt$ $0$ for $0$ $\lt$ $x$ $\lt$ $\frac{\pi}{2}$
since $\sec^{2}x$ $\gt$ $1$ for $0$ $\lt$ $x$ $\lt$ $\frac{\pi}{2}$, so $f$ is increasing on $\left(0,\frac{\pi}{2}\right)$.
Thus $f(x)$ $\gt$ $f(0)$ = $0$ for $0$ $\lt x\lt\frac{\pi}{2}$
$\tan x-x\gt0$
$\tan x\gt x$ for $0\lt x\lt \frac{\pi}{2}$