$$0 < \theta < \frac{\pi}{2}\quad\text{and}\quad 1 < n \in \mathbb{N}\quad\Rightarrow\quad \tan\theta \;>\; n\;\tan\frac{\theta}{n}$$

*Motivated by this question on the Mathematics Stack Exchange.*

Trigono**graphy** Trig, if you will, the picture ...

$$0 < \theta < \frac{\pi}{2}\quad\text{and}\quad 1 < n \in \mathbb{N}\quad\Rightarrow\quad \tan\theta \;>\; n\;\tan\frac{\theta}{n}$$

*Motivated by this question on the Mathematics Stack Exchange.*