Question

In triangle $ABC,$ prove that :
$\tan \frac{A-B}{2}=\frac{a-b}{b+a}\cot \frac{C}{2}$

Answer 1

$\begin{array}{c}In\Delta ABC\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\left(let\right)\\ \frac{a-b}{a+b}=\frac{2R\sin A-2R\sin B}{2R\sin A+2R\sin B}\\ =\frac{\sin A-\sin B}{\sin A+\sin B}\\ =\frac{2\cos \frac{A+B}{2}.\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}.\cos \frac{A-B}{2}}\\ =\cot \frac{A+B}{2}.\tan \frac{A-B}{2}\\ =\tan \frac{c}{2}.\tan \frac{A-B}{2}\\ \therefore \tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{c}{2}\end{array}$