Question

If
A,B,C are the interior angles of triangle ABC, then prove that $\text{cosec}\left(\frac{C+A}{2}\right)=\sec \left(\frac{B}{2}\right)$

Answer 1

$\because \angle A,\angle B$ and $\angle C$ are angles of a triangle

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

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

$⟹\frac{\angle A+\angle C}{2}={90}^{\circ }-\frac{\angle B}{2}$ (on dividing both sides by $2$)

$⟹\text{cosec}\left(\frac{\angle A+\angle C}{2}\right)=\text{cosec}({90}^{\circ }-\frac{\angle B}{2})$ (on applying $\text{cosec}$ both the sides)

$⟹$ $\text{cosec}\left(\frac{\angle A+\angle C}{2}\right)=\sec \left(\frac{\angle B}{2}\right)$ $[\because \text{cosec}({90}^{\circ }-\theta )=\sec \theta ]$

Hence proved.