If tan2θ = 1 - e2, then the value of secθ + tan3θ.cosecθ is?
A. $${\left( {2 + {e^2}} \right)^2}$$
B. $${\left( {2 - {e^2}} \right)^{\frac{1}{2}}}$$
C. $${\left( {2 + {e^2}} \right)^{\frac{1}{2}}}$$
D. $${\left( {2 - {e^2}} \right)^{\frac{3}{2}}}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{ta}}{{\text{n}}^2}\theta = 1 - {e^2} \cr & \therefore sec\theta + {\text{ta}}{{\text{n}}^3}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .{\text{tan}}\theta . {\text{cosec}}\theta \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .\frac{{\sin \theta }}{{{\text{cos}}\theta }}.\frac{1}{{\sin \theta }} \cr & \Rightarrow sec\theta + {\text{ta}}{{\text{n}}^2}\theta .sec\theta \cr & \Rightarrow sec\theta \left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow \sqrt {1 + {\text{ta}}{{\text{n}}^2}\theta } .\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right) \cr & \Rightarrow {\left( {1 + {\text{ta}}{{\text{n}}^2}\theta } \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {1 + 1 - {e^2}} \right)^{\frac{3}{2}}} \cr & \Rightarrow {\left( {2 - {e^2}} \right)^{\frac{3}{2}}} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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