1. **Stating the problem:** We want to find the intervals where the function $f(x) = \tan x - x$ is increasing (croissante) and decreasing (décroissante). 2. **Formula and rules:** To determine where a function is increasing or decreasing, we find its derivative $f'(x)$ and analyze its sign. - If $f'(x) > 0$, then $f$ is increasing. - If $f'(x) < 0$, then $f$ is decreasing. 3. **Find the derivative:** $$ f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(x) = \sec^2 x - 1 $$ 4. **Simplify the derivative:** Recall the identity $\sec^2 x = 1 + \tan^2 x$, so $$ f'(x) = (1 + \tan^2 x) - 1 = \tan^2 x $$ 5. **Analyze the sign of $f'(x)$:** Since $\tan^2 x \geq 0$ for all $x$ where $\tan x$ is defined, and $\tan^2 x = 0$ only when $\tan x = 0$ (i.e., $x = k\pi$, $k \in \mathbb{Z}$), we have: - $f'(x) > 0$ for all $x \neq k\pi$ where $\tan x$ is defined. - $f'(x) = 0$ at $x = k\pi$. 6. **Domain considerations:** The function $f(x) = \tan x - x$ is defined on intervals between vertical asymptotes of $\tan x$, which occur at $x = \frac{\pi}{2} + k\pi$. 7. **Conclusion:** - On each interval $\left( k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2} \right)$, $f'(x) = \tan^2 x > 0$ except at $x = k\pi$ where it is zero. - Therefore, $f$ is strictly increasing on each such interval. - There are no intervals where $f$ is decreasing. **Final answer:** The function $f(x) = \tan x - x$ is increasing on every interval $\left( k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2} \right)$ for all integers $k$, and never decreasing.