Example 5. Given the sequence $\left\{x_{0}\right\}: x_{n+1}=$ $\frac{x_{\mathrm{n}}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}$. Find the value of $x_{1001}-x_{401}$.
Given the shape of the recurrence relation, we can set $x_{\mathrm{n}}=\operatorname{tg} \alpha_{\mathrm{n}}$, and since $\operatorname{tg} \frac{\pi}{12}=2-\sqrt{3}$, we know that $x_{n+1}=\operatorname{tg} \alpha_{n+1}$
$$
=\frac{\operatorname{tg} \alpha_{\mathrm{a}}+\operatorname{tg} \frac{\pi}{12}}{1-\operatorname{tg} \alpha_{\mathrm{n}} \operatorname{tg} \frac{\pi}{12}}=\operatorname{tg}\left(\alpha_{\mathrm{n}}+\frac{\pi}{12}\right) \text {. }
$$

Therefore, $x_{\mathrm{n}+12}=x_{\mathrm{n}}$.
$$
\text { Hence } x_{1001}-x_{401}=x_{5}-x_{3}=0\left(\left\{x_{0}\right\}\right. \text { is a }
$$
periodic sequence).