Answer
$tan~x \gt x~~~$ for $~~0 \lt x \lt \frac{\pi}{2}$
Work Step by Step
Let $f(x) = tan~x-x$
$f'(x) = \frac{1}{cos^2~x}-1$
On the interval $(0, \frac{\pi}{2})$:
$0 \lt cos~x \lt 1$
$0 \lt cos^2~x \lt 1$
$\frac{1}{cos^2~x} \gt 1$
$\frac{1}{cos^2~x}-1 \gt 0$
$f'(x) \gt 0$ on the interval $(0, \frac{\pi}{2})$
$f(x) = tan~x-x~~$ is increasing on the interval $(0, \frac{\pi}{2})$.
Note that $f(0) = tan~0-0 = 0$
Therefore, on the interval $(0, \frac{\pi}{2})$:
$f(x) = tan~x-x \gt 0$
$tan~x \gt x$
$tan~x \gt x~~~$ for $~~0 \lt x \lt \frac{\pi}{2}$
