A tetrahedron in hyperbolic 3-space can be defined (up to isometry) by the measures of its dihedral angles, $(a, b, c, a^\prime, b^\prime, c^\prime)$, with $a$, $b$, $c$ along edges that meet at a vertex, and $a^\prime$, $b^\prime$, $c^\prime$ along respective opposite edges.
A Regge symmetry generates a new tetrahedron by transforming the dihedral angles of the original. One such symmetry has this effect:
$$\left(a,\frac{-b+c+b^\prime+c^\prime}{2},\frac{b-c+b^\prime+c^\prime}{2}, a^\prime, \frac{b+c-b^\prime+c^\prime}{2}, \frac{b+c+b^\prime-c^\prime}{2}\right)$$
(Regge-symmetric tetrahedra are interesting (to me) because they have the same volume. In fact, any two Regge-symmetric tetrahedra are scissors congruent: you can cut one into polyhedra that re-assemble to form the other.)
I believe it is (or may be) true that we achieve the same result by interpreting the above as a transformation of edge lengths rather than of dihedral angles.
A few bouts of arduous symbol manipulation show that this belief is consistent with a variety of properties of the dihedrally-defined Regge symmetry, but before delving into an epic verification ---fraught with many a sign/sine/sinh error--- I thought it prudent to seek-out a pointer to relevant literature.
Where may I find confirmation (or refutation) of my belief?
Note: Web searches for "Regge symmetry" ---No, Google, not "reggae symmetry"!--- reveal a number of abstract discussions of these transformations as they relate to "$6j$ symbols" and such, but (so far as I have been able to determine) none of these discussions directly addresses the geometry of the angle-edge duality. Maybe I missed one.