The function f(x) = tan x – x

The correct option is "always increases".

To determine the behavior of the function $f\left(x\right)=tanx-x$, we need to analyze its first derivative with respect to $x$.

Let the given function be:
$f\left(x\right)=tanx-x$

Differentiating both sides with respect to $x$:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}(tanx-x)$

Since the derivative of $tanx$ is ${sec}^{2}x$ and the derivative of $x$ is $1$, we get:
$f^{\text{'}}\left(x\right)={sec}^{2}x-1$

Using the fundamental trigonometric identity ${sec}^{2}x-1={tan}^{2}x$, we can rewrite the derivative as:
$f^{\text{'}}\left(x\right)={tan}^{2}x$

Since the square of any real number is always non-negative, we have:
${tan}^{2}x\ge 0$
for all $x$ in the domain of $f\left(x\right)$ (where $x\ne (2n+1)\frac{\pi }{2}$ for any integer $n$).

Furthermore, $f^{\text{'}}\left(x\right)=0$ only at isolated points where $x=n\pi$. Since the derivative is strictly positive except at these isolated points, the function $f\left(x\right)$ is strictly increasing on any interval within its domain.

Therefore, the function $f\left(x\right)=tanx–x$ always increases.