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


The law of tangent 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 tangent 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


Proof. Using the sum-to-product formulas we can rewrite the left-hand side of $(2)$ as follows


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



The formula $(2)$ reduces to the law of tangent for a triangle when $c=0$.

A related result can be found at A generalization of Mollweide's formula (rather Newton's).

Crossposted at MO.

Question: Is this generalization known?

  • 3
    $\begingroup$ I haven't seen yet such an identity but I am not a specialist of this kind of geometry. You should submit it to a journal like Journal of Classical Geometry or Crux Mathematicorum; there, the referees are likely to say whether this identity is known or not. $\endgroup$
    – Jean Marie
    Mar 17 at 7:18
  • 3
    $\begingroup$ You can also varify your identity from here :groups.io/g/Quadri-and-Poly-Geometry , in this group people may help you in varying your theorem as their research feild are related to quadrilateral and polygon. $\endgroup$
    – user999691
    Mar 17 at 10:30
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    $\begingroup$ Defining $\theta_a:=\angle ADB=\angle ACB$, you can write $a=2r\sin\theta_a$. Likewise, $b=2r\sin\theta_b$, $c=\cdots$, $d=\cdots$. Then, the expression on the right-hand side of your identity is rewritable, via sum- and difference-to-product identities, into tangents. Simple angle-chasing reduces the result to your left-hand side. Given such a straightforward derivation, I'm led to suspect that the identity exists somewhere in the literature; finding trig relations used to be a bigger deal than it is today. Of course, this does not diminish your own accomplishment in deriving it. $\endgroup$
    – Blue
    Mar 18 at 23:24
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    $\begingroup$ Hi @Blue! As the great geometer Ivan Zelich says: "it is much easier to find a direct derivation to a result that is already known." But of course, I do not rule out the possibility that you point out. $\endgroup$ Mar 19 at 0:44
  • 2
    $\begingroup$ I have no idea if this identity is well known. My guess is that it has been seen somewhere. I can provide a proof of the identity, but I am not sure that that is what you are looking for. If you have a proof and are simply looking for the identity in print, it would provide wonderful context to include the proof. $\endgroup$
    – robjohn
    Mar 19 at 19:04

1 Answer 1


The Usual Law of Tangents

Applying the formulae for the Sum of Sines and the Sum of Cosines, we get $$ \begin{align} \frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)} &=\frac{2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}{2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}\\[6pt] &=\tan\left(\tfrac{A+B}2\right)\tag1 \end{align} $$ Substituting $B\mapsto-B$ in $(1)$ and then dividing by $(1)$ gives $$ \frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag2 $$ If we let $a$ and $b$ be the sides opposite angles $A$ and $B$ respectively, the Law of Sines leads to $$ \frac{a-b}{a+b}=\frac{\sin(A)-\sin(B)}{\sin(A)+\sin(B)}\tag3 $$

Equating $(2)$ and $(3)$, we get the Law of Tangents $$ \frac{a-b}{a+b}=\frac{\tan\left(\frac{A-B}2\right)}{\tan\left(\frac{A+B}2\right)}\tag4 $$ where $a$ and $b$ are the sides of a triangle opposite angles $A$ and $B$ respectively.

Inscribed Generalization

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The power of the point $S$ is equal to both $\overline{SA}\,\overline{SD}=\overline{SB}\,\overline{SC}$; therefore, $$ \frac{\overline{SA}}{\overline{SB}}=\frac{\overline{SC}}{\overline{SD}}\tag5 $$ and since $\angle ASB=\angle CSD$, SAS says that $$ \triangle ASB\simeq\triangle CSD\tag6 $$ Therefore, $$ \frac{\overline{SA}}{\overline{SC}}=\frac{\overline{SB}}{\overline{SD}}=\frac ac\tag7 $$ Thus, we have $$ \begin{align} b&=\overline{SB}-\overline{SC}=\overline{SB}-\overline{SA}\frac ca\tag{8a}\\[6pt] d&=\overline{SA}-\overline{SD}=\overline{SA}-\overline{SB}\frac ca\tag{8b} \end{align} $$ From $(8)$, we can solve $$ \begin{align} \overline{SA}&=a\frac{ad+bc}{a^2-c^2}\tag{9a}\\[3pt] \overline{SB}&=a\frac{ab+cd}{a^2-c^2}\tag{9b} \end{align} $$ Now we are ready to apply the usual Law of Tangents from $(4)$: $$ \begin{align} \frac{\tan\left(\frac{\alpha-\beta}2\right)}{\tan\left(\frac{\alpha+\beta}2\right)} &=\frac{\overline{SB}-\overline{SA}}{\overline{SB}+\overline{SA}}\tag{10a}\\ &=\frac{(ab+cd)-(ad+bc)}{(ab+cd)+(ad+bc)}\tag{10b}\\[6pt] &=\frac{(a-c)(b-d)}{(a+c)(b+d)}\tag{10c} \end{align} $$


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