Roots of a polynomial whose coefficients are ratios of binomial coefficients 
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots +  \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$
Hence prove that
$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$

I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.
Any help will be appreciated.
Thanks.
 A: We have
$$\frac{e^{m i \theta}}{\sin^m \theta} = \frac{(\cos \theta + i \sin \theta)^m}{\sin^m \theta} = (\cot \theta + i)^m $$
for $m \in \mathbb{N}$ and $\theta \in \mathbb{R}$. Take $m=2n+1$ and $\theta = k\pi/(2n+1)$ to get
$$\frac{(-1)^k}{\sin^{2n+1}\theta} = \sum_{r=0}^{2n+1} \binom{2n+1}{r}i^r \cot^{2n+1-r} \theta . $$
The imaginary part comes from the summands for which $r$ is odd, hence
$$0 = \sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s (\cot^2\theta)^{n-s}. $$
Equivalently, the polynomial
$$\sum_{s=0}^n \binom{2n+1}{2s+1} (-1)^s x^{n-s} $$
has $\cot^2 \theta$ as a root. Divide by $2n+1$ to get the relation in the question. For the second part, take the coefficient of $x^{n-1}$ to get
$$\sum_{k=0}^n (\cot \frac{k\pi}{2n+1})^2 = \frac{1}{2n+1}\binom{2n+1}{3}. $$
Now divide by $2n(2n-1)$ and take a limit using $\cot \theta = 1/\theta + O(\theta)$.
A: Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
Multiplying the denominator & the numerator by $\cot^{2n+1}x$ we get, 
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\cot^{2n}x-\binom{2n+1}3\cot^{2n-2}x+\cdots}{\cot^{2n+1}x-\binom{2n+1}2\cot^{2n-1}x+\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is any integer
$\implies x=\dfrac{r\pi}{2n+1}$ where $r=-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$
So, the finite roots of $$\binom{2n+1}1y^{2n}-\binom{2n+1}3y^{2(n-1)}+\cdots=0$$
are $y=\cot x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=\pm1,\pm2\cdots,\pm n\pmod{2n+1}$
But $\cot(-A)=-\cot A$
So, the finite roots of $$\binom{2n+1}1z^n-\binom{2n+1}3z^{(n-1)}+\cdots=0$$
are $z=\cot^2x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=1,2\cdots,n\pmod{2n+1}$
The rest has already been taken care of in the other answer
