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, \[\tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2}\] \[\tan\left(\frac{C-A}{2}\right)=\frac{c-a}{c+a}\cot\frac{B}{2}\]

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