Question

If $A,B,C$ are the angles of a triangle, show that
$\tan \frac{A-B}{2}=\cot (\frac{C}{2}+B)$

Answer 1

We know that the sum of all angles of a triangle is 180 degree

$A+B+C={180}^{\circ }$

$\Rightarrow \tan \frac{A-B}{2}$

$=\tan \frac{{90}^{\circ }-(B+C)-B}{2}$

$=\tan {\{90}^{\circ }-(\frac{C}{2}+B)\}$

$=\cot \frac{C}{2}+B$

proved