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

Solution (By Examveda Team)

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