In a triangle ABC, sec $\dfrac{C + A}{2}$ is equal to :

1. cosec $\dfrac{B}{2}$

2. sec $\dfrac{B}{2}$

3. cosec $\dfrac{B + A}{2}$

4. none of these

In triangle ABC,

By angle sum property of triangle,

⇒ A + B + C = 180°

⇒ A + C = 180° - B

⇒ $\dfrac{A + C}{2} = \dfrac{180° - B}{2}$ ……..(1)

Substituting value of $\dfrac{A + C}{2}$ in sec $\dfrac{C + A}{2}$, we get :

$\Rightarrow \text{sec } \dfrac{C + A}{2} \\[1em] \Rightarrow \text{sec } \dfrac{180° - B}{2} \\[1em] \Rightarrow \text{sec } \Big(90° - \dfrac{B}{2}\Big) \\[1em] \Rightarrow \text{cosec } \dfrac{B}{2}.$

Hence, Option 1 is the correct option.