7. Given $\frac{\sin \theta}{\sqrt{3} \cos \theta+1}>1$, then the range of $\tan \theta$ is
Answer $\quad(-\infty,-\sqrt{2}) \cup\left(\frac{\sqrt{3}}{3}, \sqrt{2}\right)$.
Analysis From the given, we have $\frac{\sin \theta}{\cos \theta+\frac{\sqrt{3}}{3}}>\sqrt{3}$, so the slope of the line connecting point $P(\cos \theta, \sin \theta)$ and point $M\left(-\frac{\sqrt{3}}{3}, 0\right)$ is greater than $\sqrt{3}$. By plotting, we can see that point $P$ can only move on arcs $AB$ and $CD$, excluding the endpoints:
$$
A\left(-\frac{\sqrt{3}}{3}, \frac{\sqrt{6}}{3}\right), B(0,1), \quad C\left(-\frac{\sqrt{3}}{3},-\frac{\sqrt{6}}{3}\right), D\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)
$$

Therefore, $\tan \theta \in(-\infty,-\sqrt{2}) \cup\left(\frac{\sqrt{3}}{3}, \sqrt{2}\right)$.