Example 5 Given that $f(x)$ is a function defined on the set of real numbers, and $f(x+2)[1-f(x)]=1+f(x)$, if $f(5)$ $=2+\sqrt{3}$, find $f(2009)$.
$$
\begin{array}{r}
\text { Sol: } \because f(x) \neq 1, \therefore f(x+2)=\frac{1+f(x)}{1-f(x)} \\
\therefore f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} \\
=-\frac{1}{f(x)}, \therefore f(x+8)=-\frac{1}{f(x+4)}=f(x) .
\end{array}
$$

Therefore, $f(x)$ is a periodic function with a period of 8,
$$
\begin{array}{l}
\because f(5)=2+\sqrt{3}, \\
\therefore f(2009)=f(1)=-\frac{1}{f(5)}=-\frac{1}{2+\sqrt{3}}=
\end{array}
$$
$$
\sqrt{3}-2
$$