Question

If ${\tan }^2\theta =(1-e^2)$ then $\sec \theta +{\tan }^3\theta \text{cosec}\theta$ is equal to

• A. ${(2+e^2)}^{\frac{3}{2}}$
• B. ${(2-e^2)}^{\frac{3}{2}}$
• C. ${(1-e^2)}^{\frac{3}{2}}$
• D. ${(1+e^2)}^{\frac{3}{2}}$

Answer 1

The correct option is B ${(2-e^2)}^{\frac{3}{2}}$

$\sec \theta +{\tan }^3\theta \text{cosec}\theta$
$=\frac{1}{\cos \theta }+\frac{{\sin }^2\theta }{{\cos }^3\theta }=\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^3\theta }$
$={\sec }^3\theta$
Now ${\tan }^2\theta =(1-e^2)\Rightarrow {\sec }^2\theta =2-e^2$
$\therefore {\sec }^3\theta ={(2-e^2)}^{\frac{3}{2}}$