A, B, C are interior angles of ΔABC. Prove that cosec (A+B)/2 = sec C/2

### Answer

A + B + C = 180°

⇒ (A+B)/2 = 90° – C/2

⇒ cosec (A+B)/2 = cosec (90° – C/2) = sec C/2

A, B, C are interior angles of ΔABC. Prove that cosec (A+B)/2 = sec C/2

A + B + C = 180°

⇒ (A+B)/2 = 90° – C/2

⇒ cosec (A+B)/2 = cosec (90° – C/2) = sec C/2