Example 6 Given that $f(x)$ is a function with domain $R$, and $f(x+2)[1-f(x)]=1+f(x), f(1)=2+\sqrt{3}$. Find $f(1989)$.

保留源文本的换行和格式，直接输出翻译结果。
$$
\begin{array}{l}
\text { Solution: } \because f(x+2)=\frac{1+f(x)}{1-f(x)}, \\
\therefore f(x+4)=\frac{1+f(x+2)}{1-f(x+2)} \\
=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)} . \\
\text { Also, } \because f(x)=\frac{f(x+2)-1}{f(x+2)+1}, \\
\therefore f(x-4)=\frac{f(x-2)-1}{f(x-2)+1} \\
=\frac{\frac{f(x)-1}{f(x)+1}-1}{\frac{f(x)-1}{f(x)+1}+1}=-\frac{1}{f(x)} .
\end{array}
$$

Thus, $f(x+4)=f(x-4)$,
which means
$$
\begin{array}{l}
f(x+8)=f(x) . \\
\therefore f(1989)=f(248 \times 8+5) \\
\quad=f(5)=-\frac{1}{f(1)}=-\frac{1}{2+\sqrt{3}} \\
\quad=\sqrt{3}-2 .
\end{array}
$$