10. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2 \sqrt{3}, a_{n-1}=\frac{8 a_{n}}{4-a_{n}^{2}}(n \geqslant 2)$, then $a_{n}=$
10. $2 \tan \frac{\pi}{3 \cdot 2^{n-1}}$. From $a_{n-1}=\frac{8 a_{n}}{4-a_{n}^{2}}$, we get $\frac{a_{n-1}}{2}=\frac{2\left(\frac{a_{n}}{2}\right)}{1-\left(\frac{a_{n}}{2}\right)^{2}}$. Let $\frac{a_{n}}{2}=\tan \theta_{n}\left(-\frac{\pi}{2}<\theta_{n}<\frac{\pi}{2}\right)$, then $\tan \theta_{n-1}=\tan 2 \theta_{n}$. Therefore, $\theta_{n-1}=2 \theta_{n}$, and $\theta_{1}=\arctan \sqrt{3}=\frac{\pi}{3}$, so $\theta_{n}=\frac{\pi}{3} \cdot\left(\frac{1}{2}\right)^{n-1}$. Hence, $\frac{a_{n}}{2}=\tan \theta_{n}=\tan \frac{\pi}{3 \cdot 2^{n-1}}$. Therefore, $a_{n}=2 \tan \frac{\pi}{3 \cdot 2^{n-1}}$.