The function f(x) = tan x-x

The correct answer is Option (2) → is an increasing function on $[0,\frac{\pi}{2})$

$f(x)=\tan x - x$

$f'(x)=\sec^{2}x - 1$

$\sec^{2}x - 1=\tan^{2}x$

$\tan^{2}x \ge 0$ for all $x$ in $\left[0,\frac{\pi}{2}\right)$.

Thus $f'(x) > 0$ for all $x>0$, so the function is increasing on the entire interval.

Final answer: the function is increasing on $\left[0,\frac{\pi}{2}\right)$