The problem statement is incorrect: it should be $ n|k$ not $ k|n$. It is clear that if $ n|k$ then every summand is zero and so is the sum. For the rest of the proof assume that $ n\not| k$. First, notice that \[ \cot \frac{\pi m}{n}\sin \frac{2\pi km}{n} = \frac{\sin \frac{\pi (2k+1)m}{n}}{2\sin \frac{\pi m}{n}} + \frac{\sin \frac{\pi (2k-1)m}{n}}{2\sin \frac{\pi m}{n}}\] \[ =1+\sum_{j=1}^k \cos \frac{2\pi j m}{n} + \sum_{j=1}^{k-1} \cos \frac{2\pi j m}{n}=1-\cos \frac{2\pi k m}{n}+2\sum_{j=1}^k \cos \frac{2\pi j m}{n}\] and \[ \sum_{m=1}^{n-1} \cos \frac{2\pi j m}{n} = \begin{cases} -1, & n\not|j;\\ n-1,&n|j.\end{cases}\] Therefore, \[ \sum^{n-1}_{m=1}\cot \frac{\pi m}{n}\sin \frac{2\pi km}{n} = (n - 1) - (-1) + 2\left(n\left\lfloor\frac{k}{n}\right\rfloor - k\right)=n+2n\left\lfloor\frac{k}{n}\right\rfloor - 2k\] and multiplying this expression by $ -\frac{1}{2n}$ completes the proof.