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

1. 0

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

3. cosec B

4. $\text{cosec}\dfrac{B}{2}$

Given:

In Δ ABC,

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

Hence, option 4 is the correct option.