SOLUTION —
$f(x)=\tan x-x$
$\Rightarrow f^{\prime}(x)=\sec ^2 x-1=\frac{1}{\cos ^2 x}-1=\frac{1-\cos ^2 x}{\cos ^2 x}$
Since, $0 \leq \cos ^2 x \leq 1$ for all values of $x$.
$\therefore f^{\prime}(x)>0$ for all values of $x$. Thus, $f(x)$ always increases
So, The correct option will be (A).