10. Given the sequence $\left\{a_{n}\right\}, a_{1}=1, a_{n+1}$ $=\frac{\sqrt{3} a_{n}+1}{\sqrt{3}-a_{n}}$, then $\sum_{n=1}^{2022} a_{n}=$ $\qquad$
Let $a_{n}=\tan \theta_{n}$, then $\tan \theta_{n+1}=\frac{\sqrt{3} \tan \theta_{n}+1}{\sqrt{3}-\tan \theta_{n}}=\frac{\tan \theta_{n}+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3} \tan \theta_{n}}=\tan \left(\theta_{n}+\frac{\pi}{6}\right)$.
Thus, $a_{n}$ has a period of 6 and $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=0$, so $\sum_{n=1}^{2022} a_{n}=0$.