If tan2θ = 1 - e2, then the value of secθ + tan3θ.cosecθ is?

Answer: Option D

$$\begin{array}{rl}& \text{ta}{\text{n}}^2\theta =1-e^2\\ & \therefore sec\theta +\text{ta}{\text{n}}^3\theta .\text{cosec}\theta \\ & \Rightarrow sec\theta +\text{ta}{\text{n}}^2\theta .\text{tan}\theta .\text{cosec}\theta \\ & \Rightarrow sec\theta +\text{ta}{\text{n}}^2\theta .\frac{\sin \theta }{\text{cos}\theta }.\frac{1}{\sin \theta }\\ & \Rightarrow sec\theta +\text{ta}{\text{n}}^2\theta .sec\theta \\ & \Rightarrow sec\theta \left(1+\text{ta}{\text{n}}^2\theta \right)\\ & \Rightarrow \sqrt{1+\text{ta}{\text{n}}^2\theta }.\left(1+\text{ta}{\text{n}}^2\theta \right)\\ & \Rightarrow {\left(1+\text{ta}{\text{n}}^2\theta \right)}^{\frac{3}{2}}\\ & \Rightarrow {\left(1+1-e^2\right)}^{\frac{3}{2}}\\ & \Rightarrow {\left(2-e^2\right)}^{\frac{3}{2}}\end{array}$$