Question

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

Answer 1

By angle sum property of a triangle $\angle A+\angle B+\angle C={180}^{\circ }$

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

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

Now,

$\tan \left(\frac{\angle A+\angle C}{2}\right)=\tan \left(\frac{{180}^{\circ }-\angle B}{2}\right)=\tan ({90}^{\circ }-\frac{\angle B}{2})$

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

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