Question

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

Answer 1

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

Answer 2

Solution:

Given:

=> A, B, C are interior angles of ΔABC

To Prove:

$\sf{\implies cosec\Bigg(\dfrac{A+B}{2}\Bigg)=\sec \Bigg(\dfrac{C}{2}\Bigg)}$

We know that:

In ΔABC,

=> A + B + C = 180°        [Angle Sum Property]

=> A + B = 180° - C

Divide LHS and RHS part By 2,

$\sf{\implies \dfrac{A+B}{2}=\dfrac{180^{\circ}-C}{2}}$

$\sf{\implies \dfrac{A + B}{2}=90^{\circ}-\dfrac{C}{2}}$

$\sf{\implies cosec \bigg(\dfrac{A+B}{2}\bigg)=cosec \bigg(90^{\circ}-\dfrac{C}{2}\bigg)}$

We know that cosec(90° - Ф) = secФ,

$\implies {\boxed{\boxed{\bf{\red{cosec \bigg(\dfrac{A+B}{2}\bigg)=\sec \bigg(\dfrac{C}{2}\bigg)}}}}}$