TITLE: A generalization of the law of tangents
QUESTION [5 upvotes]: The law of tangents is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides.
Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle, and $\alpha$, $\beta$ and $\gamma$ be the angles opposite those three respective sides. The law of tangents states that
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{a-b}{a+b}.\tag{1}\label{1}$$
The law of tangents can be used in any case where two sides and the included angle, or two angles and a side, are known.
Although Viète gave us the modern version of the law of tangents, it was Fincke who stated the law of tangents for the first time and also demonstrated its application by solving a triangle when two sides and the included angle are given (see Wu - The Story of Mollweide and Some Trigonometric Identities).
A proof of the law of tangents is provided by Wikipedia (see here).
Generalization. Let $a$, $b$, $c$ and $d$ be the sides of a cyclic convex quadrilateral. Let $\angle{DAB}=\alpha$ and  $\angle{ABC}=\beta$, then the following identity holds
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{2}\label{2}$$
Proof. Using the sum-to-product formulas we can rewrite the left-hand side of $(2)$ as follows
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{\sin\frac12(\alpha-\beta)\cos\frac12(\alpha+\beta)}{\cos\frac12(\alpha-\beta)\sin\frac12(\alpha+\beta)}=\frac{\sin{\alpha}-\sin{\beta}}{\sin{\alpha}+\sin{\beta}}.$$
The area of a cyclic quadrilateral can be expressed as $\Delta=\frac12(ad+bc)\sin{\alpha}$ (see $(12)$ at Killing three birds with one stone) and similarly for the other angles. Then substituting, simplifying and factorizing we have
$$\begin{align*}\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}&=\frac{\frac{2\Delta}{ad+bc}-\frac{2\Delta}{ab+cd}}{\frac{2\Delta}{ad+bc}+\frac{2\Delta}{ab+cd}}=\frac{ab-ad+cd-bc}{ab+ad+cd+bc}=\frac{(a-c)(b-d)}{(a+c)(b+d)}\end{align*}.$$
$\square$
The formula \eqref{2} reduces to the law of tangents for a triangle when $c=0$.
A related result can be found at A generalization of Mollweide's Formula (rather Newton's).
Crossposted at MathSE.
Question: Is this generalization known?

REPLY [7 votes]: To place the formula of the OP into context, it is helpful to note the identity (from Josefsson - More characterizations of cyclic quadrilaterals)
$$\frac{q-p}{q+p}=\frac{(a-c)(b-d)}{(a+c)(b+d)}.\tag{$*$}\label{star}$$
Here $q$ and $p$ are the lengths of the blue and red diagonals, with opposite angles $\alpha$ and $\beta$.

The quadrilateral tangent formula can thus be rewritten in a form that looks more like the triangle tangent formula,
$$\frac{\tan\frac12(\alpha-\beta)}{\tan\frac12(\alpha+\beta)}=\frac{q-p}{q+p}.$$
Q: Is it obvious that the law of tangents applies to the blue and red diagonals?


The formula \eqref{star} is equivalent (upon rearrangement of the terms) to what is known as Ptolemy’s second theorem
$$\frac{p}{q}=\frac{ad+bc}{ab+cd}.$$