If $a_k:=\tan( \sqrt{2} + \frac{k\pi}{2011})$, then evaluate $ \frac{a_1+a_2+\cdots+a_{2011}}{a_1a_2~\cdots~a_{2011}}$ 
Set $$
a_k = \tan{\left( \sqrt{2} + \frac{ k \pi   }{2011}  \right)}\quad\text{for}~k=1,2,3,\ldots, 2011
$$
  Find the value of
  $$
\frac{a_1 + a_2 + \ldots + a_{2011}}{a_1 a_2 \ldots a_{2011}}
$$

Can anyone help me in this? It's going too long.

Using Sum of tangent functions where arguments are in specific arithmetic series,
$$\tan(2m+1)x=\dfrac{\binom{2m+1}1 t-\binom{2m+1}3t^3+\cdots+(-1)^m\binom{2m+1}{2m+1}t^{2m+1}}{\binom{2m+1}0-\binom{2m+1}2t^2+\cdots++(-1)^m\binom{2m+1}{2m}t^{2m}}$$
So, if $\tan(2m+1)x=\tan y$
$(2m+1)x=k\pi+y$ where $k$ is any integer
$x=\dfrac{k\pi+y}{2m+1}; k=1,2,\cdots,2m+1$
So, the roots of $$\tan y=\dfrac{\binom{2m+1}1 t-\binom{2m+1}3t^3+\cdots+(-1)^m\binom{2m+1}{2m+1}t^{2m+1}}{\binom{2m+1}0-\binom{2m+1}2t^2+\cdots+(-1)^m\binom{2m+1}{2m}t^{2m}}$$
$$\iff(-1)^mt^{2m+1}-\tan y(-1)^m(2m+1)t^{2m}+\cdots+\tan y=0$$ are $\tan x;x=\dfrac{k\pi+y}{2m+1}, k=1,2,\cdots,2m+1$
Can apply Vieta's formula?
