Question

In $△ABC$, prove that $\tan \left(\frac{A+B}{2}\right)=\cot \frac{C}{2}$.

Answer 1

In any traingle $ABC,\angle A+\angle B+\angle C={180}^{\circ }$

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

$\Rightarrow \frac{\angle A+\angle B}{2}=\frac{{180}^{\circ }-\angle C}{2}$

$\Rightarrow \frac{\angle A+\angle B}{2}={90}^{\circ }-\frac{\angle C}{2}$

$\Rightarrow \tan \left(\frac{\angle A+\angle B}{2}\right)=\tan ({90}^{\circ }-\frac{\angle C}{2})$

$\therefore \tan \left(\frac{\angle A+\angle B}{2}\right)=\cot \frac{\angle C}{2}$

Hence proved.