Law of Tangents

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

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

From the Law of Sines, let:
 * $d = \dfrac a {\sin A} = \dfrac b {\sin B}$

so that: