To place the formula of the OP into context, it is helpful to note the identity (from <A HREF="https://ijgeometry.com/wp-content/uploads/2019/09/14-32.pdf">Josefsson - More characterizations of cyclic quadrilaterals</A>)

$$\frac{q-p}{q+p}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{$*$}\label{star}$$
Here $q$ and $p$ are the lengths of the blue and red diagonals, with opposite angles $\alpha$ and $\beta$.

<IMG SRC="https://ilorentz.org/beenakker/MO/quadrilateral.jpg" WIDTH="250" ALT="Inscribed quadrilateral, with labelled angles 𝛼 and 𝛽 and diagonals p and q"/>

The quadrilateral tangent formula can thus be rewritten in a form that looks more like the <A HREF="https://en.wikipedia.org/wiki/Law_of_tangents">triangle tangent formula</A>,
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{q-p}{q+p}.$$

**Q:** *Is it obvious that the law of tangents applies to the blue and red diagonals?*

---

<sub>
The formula \eqref{star} is equivalent (upon rearrangement of the terms) to what is known as <A HREF="https://jcgeometry.org/Articles/Volume4/Krishna.pdf">Ptolemy’s second theorem</A>
$$\frac{p}{q}=\frac{ad+bc}{ab+cd}.$$
</sub>