Solution :
According to question
$A+B+C= 180^{\circ}$
=> $ \large\frac{A+B+C}{2} $$=90^{\circ}$
=> $ \large\frac{A}{2}+\frac{B+C}{2} =90^{\circ}$
=> $ \tan \bigg( \large\frac{B+C}{2} \bigg) =$$ \tan \bigg(90^{\circ} - \frac{A}{2}\bigg)$
=> $\tan \bigg( \large\frac{B+C}{2}\bigg) =\cot \large\frac{A}{2}$