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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?

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?

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?

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?

It might be of interest to note a related formula (from Josefsson - More characterizations of cyclic quadrilaterals)

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

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 blue and red diagonals, with opposite angles $\alpha$ and $\beta$.

Is it obvious that the law of tangents applies to these two diagonals and their opposite angles?

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?