The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$, \begin{align*} a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}. \end{align*}What is $a_{2009}$?
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The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$, \begin{align*} a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}. \end{align*}What is $a_{2009}$?
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The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$, \begin{align*} a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}. \end{align*}What is $a_{2009}$?
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