Let us consider rational function $A(x,n)$ ($n\in\mathbb{N}$, $x\in D=\mathbb{R}\setminus \{-1,1\}$): $$A(x,n)=\det\begin{pmatrix} 1 & [1]&0&0& \ldots & 0 &0\\ x^{1\times2} & [n-1] &[2]&0& \ldots & 0&0\\ x^{2\times3} & 0 &[n-2]&[3]& \ldots & 0&0\\ \vdots &\vdots &\vdots & \ddots & \ddots & \vdots&\vdots &\\ x^{(n-2)(n-1)} & 0&0& 0 &\ldots & [2]&[n-1]\\ x^{(n-1)n} & 0&0& 0 &\ldots & 0&[1] \end{pmatrix}$$ where $[k] = \frac{1}{x^{2k}-1}$ and all coefficients belong to $\mathbb{R}$.
It seems to be eerie function, but I have a hypothesis (I checked it for $1\le n\le 8$) that for all $n$ and $x\in D$ we actually have $A(x,n)=(-1)^{n+1}$.
I tried to prove it but I failed.
One can use Laplace's formula to reduce the calculation much simpler.
$\det A(x,n)=1\cdot \det A(x,n)_{1,1}-x^{1\cdot 2}\det A(x,n)_{2,1}+\cdots$.
Here $A(x,n)_{i,j}$ means the $(i,j)$-th minor of the matrix $A(x,n)$.
And each of the minors $A(x,n)_{i,1}$ is of 2-block-diagonal form and the two blocks are triangular.
So we can only extract and multiply the diagonal terms to compute the determinant of each minor.
Then the result is $\det A(x,n)_{i,1}=[1][2]\cdots [i-1]\cdot [n-i]\cdots [1]$.
What is left is just a simple high-school algebra.
FYI, I want to add the remaining details and its relation with q-binomials.
So, we have the formula
$ \det A(x,n)=\sum_{i=1}^n (-1)^{i-1}x^{(i-1)i}[1]\cdots [i-1]\cdot [n-i]\cdots [1]$ $=[n-1]\cdots [1] \sum_{i=1}^n (-1)^{i-1}x^{(i-1)i} \frac{[1]\cdots [i-1]}{[n-1]\cdots [n-i+1]}$.
The summands in the last expression are called q-binomials $\binom{n-1}{i-1}_q$, for $q=x^2$ in this case.
So, ignoring the factor $[n-1]\cdots [1]$, what we want to show is the q-binomial theorem $\sum_{i=0}^{n-1}(-1)^i q^{\frac{(i-1)i}{2}}\binom{n-1}{i}_q=\prod_{i=0}^{n-1}(1-q^i)$.