Question

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

Answer 1

Given $A,B,C$ are the interior angles of $△ABC$

$⟹A+B+C={180}^{\circ }$

$⟹A+B={180}^{\circ }-C$

$LHS=\text{cosec}(\frac{A+B}{2})$

$=\text{cosec}(\frac{{180}^{\circ }-C}{2})$

$=\text{cosec}({90}^{\circ }-\frac{C}{2})$

$=\text{sec}\frac{C}{2}$

$=RHS$