Law of Tangents

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Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

$\dfrac {a + b} {a - b} = \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }$


Corollary

$\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$


Proof

Let $d = \dfrac a {\sin A}$.

From the Law of Sines, let:

$d = \dfrac a {\sin A} = \dfrac b {\sin B}$

so that:

\(\ds a\) \(=\) \(\ds d \sin A\)
\(\ds b\) \(=\) \(\ds d \sin B\)
\(\ds \leadsto \ \ \) \(\ds \frac {a + b} {a - b}\) \(=\) \(\ds \frac {d \sin A + d \sin B} {d \sin A - d \sin B}\)
\(\ds \) \(=\) \(\ds \frac {\sin A + \sin B} {\sin A - \sin B}\)
Let $C = \frac 1 2 \paren {A + B}$ and $D = \frac 1 2 \paren {A - B}$, and proceed as follows:
\(\ds \leadsto \ \ \) \(\ds \frac {a + b} {a - b}\) \(=\) \(\ds \frac {2 \sin C \cos D} {\sin A - \sin B}\) Sine plus Sine
\(\ds \) \(=\) \(\ds \frac {2 \sin C \cos D} {2 \sin D \cos C}\) Sine minus Sine
\(\ds \) \(=\) \(\ds \frac {\frac {\sin C} {\cos C} } {\frac {\sin D} {\cos D} }\) dividing top and bottom by $\cos C \cos D$
\(\ds \) \(=\) \(\ds \frac {\tan C} {\tan D}\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }\) substituting back for $C$ and $D$

$\blacksquare$


Also see


Sources

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