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

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

The tangent formula in the OP can thus alternatively be written as
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{q-p}{q+p},$$
where $q$ and $p$ are the lengths of the red and blue diagonals, with opposite angles $\alpha$ and $\beta$.

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