This is a sketch not a proof, but if it's not enough let me know:

The objective is to construct a poly whose roots are

${\cot }^2(\pi k/(2m+1))$

If the poly is monic, then the second highest coefficient will be (to within a minus sign) the sum of the roots, i.e. your expression.

How do we construct the polynomial? Consider (for odd $p$)

$(\cos \varphi +i\sin \varphi {)}^p=(\cos p\varphi +i\sin p\varphi )$

$=\langle \mathrm{power}\mathrm{series}\mathrm{in}\mathrm{products}\mathrm{of}$, $\rangle$

Take the imaginary part (I'll leave you to figure out the coefficients $a$, $b$, ...)

$\sin p\varphi =a{\sin }^p\varphi +b{\sin }^{p-2}\varphi {\cos }^2\varphi +...$

Now take out a factor of ${\sin }^p\varphi$ from the RHS (I'll leave you to figure out the $m$, $n$...)

$\sin p\varphi ={\sin }^p\varphi (m{\cot }^{p-1}\varphi +n{\cot }^{p-3}\varphi +...)$

Now we know (from the LHS) the $p\text{'}$ for which the RHS is zero. So letting $z={\cot }^2\varphi$ the RHS is a poly in $z$, the roots of which are ${\cot }^{2}p\text{'}$.

So the sum of the roots is $-n/m$. You now need to figure out what to choose for $p$, $\varphi$ and what $m$, $n$ are.

Let me know if this is not enough.

I don't know whether you managed to solve the problem, so let me go through a particular example ( $m=2$) using the approach outlined above.

Consider

$(\cos (\mathrm{theta})+i\sin (\theta {))}^5=\cos (5\theta )+i\sin (5\theta )$

$=1(\cos (\theta {))}^5(i\sin (\theta {))}^0+5(\cos (\theta {))}^4(i\sin (\theta {))}^1+10(\cos (\theta {))}^3(i\sin (\theta {))}^2+10(\cos (\theta {))}^2(i\sin (\theta {))}^3+5(\cos (\theta {))}^1(i\sin (\theta {))}^4+1(\cos (\theta {))}^0(i\sin (\theta {))}^5$

Take imaginary part:

$\sin (5\theta )=5(\cos (\theta {))}^4(\sin (\theta {))}^1-10(\cos (\mathrm{theta}{))}^2(\sin (\theta {))}^3+1(\cos (\theta {))}^0(\sin (\theta {))}^5$

Taking out ${\sin }^5(\theta )$ from the RHS:

$\sin (5\theta )={\sin }^5(\theta )[5{\cot }^4(\theta )-10{\cot }^2(\theta )+1]$

The LHS is zero when $\theta =0$, $\pi /5$, $2\pi /5$, $3\pi /5$, $4\pi /5$, $\pi$, etc. For the RHS, $\sin (\theta )$ is zero when $\theta =0$, $\pi$ etc. Therefore

$5{\cot }^4(\theta )-10{\cot }^2(\theta )+1$

is zero when $\theta =\pi /5$, $2\pi /5$, $3\pi /5$, $4\pi /5$, $6\pi /5$ etc.

Write $z={\cot }^2(\theta )$, the equation

$5x^2-10z+1=0$ has distinct roots

$z={\cot }^2(\pi /5)$, ${\cot }^2(2\pi /5)$

But the sum of the roots of this quadratic are $10/5=2$, hence ${\cot }^2(\pi /5)+{\cot }^2(2\pi /5)=2$

I'll leave you to generalise it for any $m$, if you haven't already done so.