Note the similarity of the recursion to the angle addition identity \[\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}.\]We can take advantage of this similarity as follows: Let $ f_1 = 3, $ $ f_2 = 2, $ and let $ f_n = f_{n - 1} + f_{n - 2} $ for all $ n \ge 3 $. Let $ \theta_n = \frac{f_n \pi}{12} $. Then $ \tan \theta_1 = \tan \frac{\pi}{4} = 1 $ and $ \tan \theta_2 = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} $. Also, \begin{align*} \tan \theta_{n + 2} &= \tan (\theta_{n + 1} + \theta_n) \\ &= \frac{\tan \theta_{n + 1} + \tan \theta_n}{1 - \tan \theta_n \tan \theta_{n + 1}}.\end{align*}Since the sequences $ (a_n) $ and $ (\tan \theta_n) $ have the same initial terms, and the same recursion, they coincide. Since $ \tan \theta $ is periodic with period $ \pi, $ to compute further terms of $ \tan \theta_n, $ it suffices to compute $ f_n $ modulo 12: \[ \begin{array}{c|c} n & f_n \pmod{12} \\ \hline 1 & 3 \\ 2 & 2 \\ 3 & 5 \\ 4 & 7 \\ 5 & 0 \\ 6 & 7 \\ 7 & 7 \\ 8 & 2 \\ 9 & 9 \\ 10 & 11 \\ 11 & 8 \\ 12 & 7 \\ 13 & 3 \\ 14 & 10 \\ 15 & 1 \\ 16 & 11 \\ 17 & 0 \\ 18 & 11 \\ 19 & 11 \\ 20 & 10 \\ 21 & 9 \\ 22 & 7 \\ 23 & 4 \\ 24 & 11 \\ 25 & 3 \\ 26 & 2 \end{array} \]Since $ a_{25} \equiv a_1 \pmod{12} $ and $ a_{26} \equiv a_2 \pmod{12}, $ the sequence modulo 12 become periodic at this point, with period 12. Therefore, \[a_{2009} = \tan \theta_{2009} = \tan \theta_5 = \boxed{0}.\]