Cotangent summation (proof) How to sum up this thing, i tried it with complex number getting nowhere so please help me with this,$$\sum_{k=0}^{n-1}\cot\left(x+\frac{k\pi}{n}\right)=n\cot(nx)$$ 

Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\cot(nx)=\dfrac1{\tan(nx)}=\dfrac{1-\binom n2\tan^2x+\cdots}{\binom n1\tan x-\binom n3\tan^3x+\cdots}=\dfrac{\cot^nx-\binom n2\cot^{n-2}x+\cdots}{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}(\text{multiplying the N & D by }\cot^nx)$$
If $\cot(nx)=\cot(nA)\iff\tan nx=\tan nA, nx=nA+m\pi$ where $m$ is any integer
$x=A+\dfrac{m\pi}n$ where $m\equiv0,1,\cdots,n-2,n-1\pmod n$
So, the roots of $$\cot^nx-\binom n1\cot nA\cdot\cot^{n-1}x-\binom n2\cot^{n-2}x+\cdots=0$$  are $\cot\left(A+\dfrac{m\pi}n\right)$ where $m\equiv0,1,\cdots,n-2,n-1\pmod n$
$$\implies\sum_{m=0}^{n-1}\cot\left(A+\dfrac{m\pi}n\right)=\binom n1\cot nA=n\cot nx$$
