Question

$A,B,C$ are interior angles of $△ABC$. Prove that $\csc \left(\frac{A+B}{2}\right)=\sec \frac{C}{2}$.

Answer 1

$A,B,C$ are interior angles of $△ABC$

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

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

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

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

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

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

Hence proved.