The function f(x)=tan x -x for x∈ ≤ft(0, (π)/2) is :

The correct answer is Option (1) → increase function

$f(x)=tan x -x $

so $f'(x)=\sec x-1$ for $ x\in \left(0, \frac{\pi}{2}\right)$

$\sec x>1$

so $f'(x)>0$

$f(x)$ → increasing