Seeking elegant proof of $\sum_{cyc}\cot(x-y)\cot(y-z)=1$ Problem
The following identity is obvious:
$$
\frac{1}{(x-y)(y-z)} + \frac{1}{(y-z)(z-x)} + \frac{1}{(z-x)(x-y)} = 0
$$
This post is for a trigonometric version in terms of cotangent:
$$
\cot (x- y) \cot( y- z) + \cot( y-z) \cot( z - x) + \cot ( z- x) \cot ( x-y) = 1
$$
The following Mathematica code gives me 1
Simplify[Cot[x - y] Cot[y - z] + Cot[z - x] Cot[y - z] +  Cot[x - y] Cot[z - x]]

So I think the identity is right.
I could have proved it using a brute force expansion in terms of trigonometric functions of $x, y,z$ individually.
But I feel there definitely is a more elegant proof. I'm looking for such a proof.
Follow-up question: is there a similar identity involving the elliptic function sn, cn, dn? I don't know the answer, but it is possible based on the context of the real world problem.
Thanks!
 A: Given any three numbers $\;x,y,z\;$ define
$$ X := e^{ix},\; Y := e^{iy},\; := e^{iz}\quad
\text{ and }\quad a := X^2,\; b := Y^2,\; c := Z^2. $$
Then, by definition of cotangent
$\; \cot(x-y) = i(a+b)/(a-b).\;$ Thus
$$ \cot(x-y)\cot(y-z) = -\frac{(a+b)(b+c)}{(a-b)(b-c)}. $$
Cycle, add, and use common denominator to get
$$  1 - \sum_{cyc}\cot(x-y)\cot(y-z)  = \frac{S}{ (a-b)(b-c)(c-a)} $$
where $$ S := (a\!-\!b)(b\!-\!c)(c\!-\!a) \!+\!
  (a\!+\!b)(b\!+\!c)(c\!-\!a)\!+\!
   (a\!-\!b)(b\!+\!c)(c\!+\!a)\!+\!
  (a\!+\!b)(b\!-\!c)(c\!+\!a) .$$
Now $\;S\;$ expands to identically zero. There are several ways
to verify this. For example, if
$$ T(b) := (a-b)(b-c)+(a+b)(b+c) = (b+b)(c+a),$$
(which is equivalent to id$_{3,3,1,2e}$) then $S = T(b)(c-a) + T(-b)(c+a), $ but $ T(b)=-T(-b). $
Thus, $\;0 = S\;$ is the identity named as id$_{3,4,1,3b}$
in my collection of hundreds of Special Algebraic Identities. This answers
your main question.
You also asked

Follow-up question: is there a similar identity involving the elliptic function sn, cn, dn? I don't know the answer, but it is possible based on the context of the real world problem.

Depends on what precisely you mean by "similar identity". I suggest that you
look at my identities for those tagged [JE] just as id$_{3,4,1,3b}$. That
is, this identity is already satisfied by Jacobi sn, sc, sd. More precisely,
suppose the function $\;f\;$ is any one of $\;\sin,\;\tan,\;$sn, sc, sd.
Then
$$ 0 = f(a-b)f(b-c)f(c-a) + f(a+b)f(b+c)f(c-a) + \\
   f(a-b)f(b+c)f(c+a) +  f(a+b)f(b-c)f(c+a).  $$
There are several other identities labeled [JE]
that all have the same solutions as this identity.
This is related to
MSE question 4088531
about functional equations for both sine and tangent.
