It might be of interest to note a related formula (from <A HREF="https://ijgeometry.com/wp-content/uploads/2019/09/14-32.pdf">Josefsson - More characterizations of cyclic quadrilaterals</A>)

$$\frac{q-p}{q+p}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{$*$}\label{star}$$

The quadrilateral tangent formula in the OP can thus be rewritten in a form that looks more like the triangle tangent formula,
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{q-p}{q+p},$$
where $q$ and $p$ are the lengths of the red and blue diagonals, with opposite angles $\alpha$ and $\beta$.

<IMG SRC="https://i.sstatic.net/nSvfc.jpg" WIDTH="250" ALT="Inscribed quadrilateral, with labelled angles 𝛼 and 𝛽 and diagonals p and q"/>

The formula \eqref{star} follows by rewriting "Ptolemy’s second theorem" 
$$\frac{p}{q}=\frac{ad+bc}{ab+cd}.\tag{$**$}$$