I've been looking at some graph polynomials and I found a very nice relation between the famous Tutte polynomial of graphs and the no less famous Jones polynomial of links.
Using this relation I was able to show, that for an alternating link $ L $ with an alternating diagram $D$:
$$ \lvert V_{L} (-1) \rvert = \# \{ \text{spanning trees of the Tait graph of } D \}. $$
(I used the Tutte polynomial of the Tait graph of $D$.)
Then I found in this paper, the equality: $$ \lvert V_{L} (-1) \rvert = \det (L) $$ for $ L $ an alternating link.
So my question is:
For an alternating link $ L $ with an alternating diagram $ D $, how do I prove that $$ \det (L) = \# \{ \text{spanning trees of the Tait graph of } D \}? $$
Thank you in advance for your help.
I think the equality you have shown is well known, though I don't know where its written down (this is somewhat close but orthogonal to my interests so don't take this opinion too seriously). Actually, I think the first author of the paper you linked mentioned it in a recent talk I attended.
I am not sure what you want to show if you already believe the equalities you've written down, but in fact $V_L(-1)=\Delta_L(-1)$ is true for any link (where $\Delta_L(t)$ is the Alexander polynomial and the determinant is usually defined as $|\Delta_L(-1)|$).
According to Wolfram.Mathworld, the equality $V_L(-1)=\Delta_L(-1)$ is present in Jones' 1985 paper where he introduced the polynomial, so that might be a good place to start.
For entertainment purposes: there is an "interesting" interpretation to the relationship you proved by a quite famous mathematician outside of knot theory http://www.math.rutgers.edu/~zeilberg/Opinion1.html