Prove:
$$\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}...\cot \frac{(m-1)\pi}{2m}=1$$
This is a roots of unity problem. I managed to show a similar example for $\cos$ by the following:
$z^m-1=0$ with roots $1,e^{2\pi /m},...,e^{2(m-1)\pi /m}$ Put the roots into a polynomial with linear factors:
$(z-1)(z-e^{2\pi /m})...(e^{2(m-1)\pi /m})=z^m-1$ and divide by $z-1$ to get:
$$(z-e^{2\pi /m})...(e^{2(m-1)\pi /m})=1+z+...+z^{m-1}$$
Evaluation at $z=1$ with some complex conjugation (I multiply 2 conjugated equations) yields the result.
I'll be happy to expand on my current work if required.
Edit: Thought I'd add a similar $\sin$ summation
We use the fact that $$\left(\cot x\right)\left(\cot\left(\frac{\pi}{2}-x\right)\right)=1.$$ The product of the entry that is $k$ from the beginning and the entry that is $k$ from the end is $1$. (If $m$ is even, there is a "middle" term, but it is $1$.)