# Difference between revisions of "Law of Tangents"

The Law of Tangents is a rather obscure trigonometric identity that is sometimes used in place of its better-known counterparts, the law of sines and law of cosines, to calculate angles or sides in a triangle.

## Statement

If $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then $$\frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} .$$

## Proof

Let $s$ and $d$ denote $(A+B)/2$, $(A-B)/2$, respectively. By the Law of Sines, $$\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .$$ By the angle addition identities, $$\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan \frac{1}{2} (A-B)}{\tan \frac{1}{2} (A+B)}$$ as desired. $\square$

## Problems

### Intermediate

In $\triangle ABC$, let $D$ be a point in $BC$ such that $AD$ bisects $\angle A$. Given that $AD=6,BD=4$, and $DC=3$, find $AB$.

Show that $[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$.