$\dfrac{\text{tan θ}}{\text{sec θ - 1}} + \dfrac{\text{tan θ}}{\text{sec θ + 1}}$ is equal to :

1. 2 cosec θ

2. 2 sec θ

3. 2 tan θ

4. sec θ tan θ

Solving,

$\Rightarrow \dfrac{\text{tan θ}}{\text{sec θ - 1}} + \dfrac{\text{tan θ}}{\text{sec θ + 1}} \\[1em] \Rightarrow \dfrac{\text{tan θ(sec θ + 1) + \text{tan θ(sec θ - 1)}}}{\text{(sec θ - 1)(sec θ + 1)}} \\[1em] \Rightarrow \dfrac{\text{tan θ. sec θ + tan θ + tan θ.sec θ - tan θ}}{\text{sec}^2 θ - 1} \\[1em] \Rightarrow \dfrac{\text{2 tan θ. sec θ}}{\text{ tan }^2 θ} \\[1em] \Rightarrow \dfrac{\text{2 sec θ}}{\text{tan θ}} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\text{cos θ}}}{\dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\text{cos θ}} \times \text{cos θ}}{\text{sin θ}} \\[1em] \Rightarrow \dfrac{2}{\text{sin θ}} \\[1em] \Rightarrow \text{2 cosec θ}.$

Hence, Option 1 is the correct option.