The Law of Tangents

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created March 26, 2026

Given a triangle \( \bigtriangleup ABC \) with \( \angle A = \alpha, \angle B = \beta, \angle C = \gamma, \overline{BC}=a, \overline{CA}=b \) and \( \overline{AB}= c \), the Law of Tangents is: $$ \begin{eqnarray*} \frac{a-b}{a+b} &=& \frac{\tan\left( \frac{\alpha-\beta}{2} \right)} {\tan\left( \frac{\alpha+\beta}{2} \right)}, \\ \frac{b-c}{b+c} &=& \frac{\tan\left( \frac{\beta-\gamma}{2} \right)} {\tan\left( \frac{\beta+\gamma}{2} \right)}, \\ \frac{c-a}{c+a} &=& \frac{\tan\left( \frac{\gamma-\alpha}{2} \right)} {\tan\left( \frac{\gamma+\alpha}{2} \right)}. \end{eqnarray*} $$ Because \( \tan(-\theta) = -\tan(\theta) \), switching the order yields another set of formulas: $$ \begin{eqnarray*} \frac{b-a}{b+a} &=& \frac{\tan\left( \frac{\beta-\alpha}{2} \right)} {\tan\left( \frac{\beta+\alpha}{2} \right)}, \\ \frac{c-b}{c+b} &=& \frac{\tan\left( \frac{\gamma-\beta}{2} \right)} {\tan\left( \frac{\gamma+\beta}{2} \right)}, \\ \frac{a-c}{a+c} &=& \frac{\tan\left( \frac{\alpha-\gamma}{2} \right)} {\tan\left( \frac{\alpha+\gamma}{2} \right)}. \end{eqnarray*} $$ The Law of Tangents can easily be proved by Mollweide's Formulas. The Mollweide's Formulas are: $$ \begin{eqnarray*} \frac{a-b}{c} &=& \frac{\sin\left( \frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\gamma}{2}\right)} \\ \frac{a+b}{c} &=& \frac{\cos\left( \frac{\alpha-\beta}{2}\right)}{\sin\left( \frac{\gamma}{2}\right)}. \end{eqnarray*} $$ Dividing the first Mollweide's formula by the second almost yields the desired result: \[ \frac{a-b}{a+b} = \frac{\left( \frac{\sin\left( \frac{\alpha-\beta}{2}\right) }{\cos\left( \frac{\gamma}{2} \right)} \right) }{\left( \frac{\cos\left( \frac{\alpha-\beta}{2}\right) }{\sin\left( \frac{\gamma}{2} \right)}\right)} = \frac{\sin\left( \frac{\alpha-\beta}{2}\right)}{\cos\left( \frac{\alpha-\beta}{2}\right)} \cdot \frac{\sin\left( \frac{\gamma}{2} \right)}{\cos\left( \frac{\gamma}{2} \right)} = \tan\left( \frac{\alpha-\beta}{2}\right)\cdot \tan\left( \frac{\gamma}{2} \right)\]

Because \( \frac{\gamma}{2} = \frac{180^{\circ}-(\alpha+\beta)}{2} = 90^{\circ} - \frac{\alpha+\beta}{2} \), we have \[ \tan\left( \frac{\gamma}{2} \right) = \frac{\sin\left( 90^{\circ} - \frac{\alpha+\beta}{2}\right)} {\cos\left( 90^{\circ} - \frac{\alpha+\beta}{2}\right)} = \frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\sin\left(\frac{\alpha+\beta}{2}\right)} = \frac{1}{\tan\left(\frac{\alpha+\beta}{2}\right)}. \] Finally, we have the desired result: \[ \frac{a-b}{a+b} = \frac{\tan\left( \frac{\alpha-\beta}{2}\right)}{\tan\left(\frac{\alpha+\beta}{2}\right)}. \]

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