Example 6 Given that the function is defined on the domain $\mathbf{R}$, and $f(x+2)[1-f(x)]=1+f(x), f(1)=$ $2+\sqrt{2}$. Find $f(1989)$.
(1989 Beijing Competition Question)
Given $f(x+2)[1-f(x)]=1+f(x)$, we have $f(x+2)=\frac{1+f(x)}{1-f(x)}$.
Thus, $f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)}$.
Also, $f(x)=\frac{f(x+2)-1}{f(x+2)+1}$, so
$f(x-4)=\frac{f(x-2)-1}{f(x-2)+1}=\frac{\frac{f(x)-1}{f(x)+1}-1}{\frac{f(x)-1}{f(x)+1}+1}=-\frac{1}{f(x)}$.
Therefore, $f(x+4)=f(x-4)$,
which means $f(x+8)=f(x)$.
Thus, $f(1989)=f(248 \cdot 8+5)=f(5)=-\frac{1}{f(1)}=-\frac{1}{2+\sqrt{3}}=\sqrt{3}-2$.