Triangle

# The Tangent Law

The tangent law states that in any triangle $ABC$, $\tan\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot\frac{C}{2}$ $\tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2}$ $\tan\left(\frac{C-A}{2}\right)=\frac{c-a}{c+a}\cot\frac{B}{2}$

Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively.

If $R$ be the circum-radius of the triangle $ABC$, then from sine law, we have, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\therefore a=2R\sin A$ $\therefore b=2R\sin B$ $\therefore c=2R\sin C$

Now, $\frac{a-b}{a+b}=\frac{2R\sin A-2R\sin B}{2R\sin A+2R\sin B}$ $=\frac{\sin A-\sin B}{\sin A+\sin B}$ $=\frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$

In any $\Delta ABC$, $A+B+C=π$ $\therefore\frac{A+B}{2}=\frac{π}{2}-\frac{C}{2}$

$\therefore\frac{a-b}{a+b}=\frac{\cos\left(\frac{π}{2}-\frac{C}{2}\right)\sin\left(\frac{A-B}{2}\right)}{\sin\left(\frac{π}{2}-\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)}$ $=\frac{\cos\frac{C}{2}\sin\left(\frac{A-B}{2}\right)}{\sin\frac{C}{2}\cos\left(\frac{A-B}{2}\right)}$ $=\tan\frac{A}{2}\tan\left(\frac{A-B}{2}\right)$ $\therefore\tan\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot\frac{C}{2}$

Similarly, we can prove that, $\tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2}$ $\tan\left(\frac{C-A}{2}\right)=\frac{c-a}{c+a}\cot\frac{B}{2}$