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

Given:

\( A, B, C \), are the interior angles of a triangle \( A B C \).

To do:

We have to prove that \( \tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2} \).

Solution:  

We know that,

$tan\ (90^{\circ}- \theta) = cot\ \theta$

Sum of the angles in a triangle is $180^{\circ}$.

This implies,

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

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

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

Therefore,

$\tan \left(\frac{B+C}{2}\right)=\tan (\frac{B}{2}+\frac{C}{2})$

$=\tan (90^{\circ}-\frac{A}{2})$

$=\cot \frac{A}{2}$

Hence proved.