Example 1 Given the sequence $\left\{x_{n}\right\}$, and
$$
x_{n+1}=\frac{x_{n}+(2-\sqrt{3})}{1-(2-\sqrt{3}) x_{n}} \text {. }
$$

Find the value of $x_{1001}-x_{401}$.
Considering the similarity between equation (1) and
$$
\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}
$$

we can make the trigonometric substitution $x_{n}=\tan \alpha_{n}$.
Also, $\tan 15^{\circ}=2-\sqrt{3}$, then
$$
\begin{array}{l}
x_{n+1}=\frac{\tan \alpha_{n}+\tan 15^{\circ}}{1-\tan \alpha_{n} \cdot \tan 15^{\circ}}=\tan \left(\alpha_{n}+15^{\circ}\right), \\
x_{n+12}=\tan \left(\alpha_{n}+12 \times 15^{\circ}\right)=\tan \alpha_{n}=x_{n} .
\end{array}
$$

Therefore, $\left\{x_{n}\right\}$ is a periodic sequence with a period of 12.
Thus, $x_{1001}-x_{401}=0$.