Question

If ${\tan }^2\theta =(1-e^2)$, prove that
$\sec \theta +{\tan }^3\theta \csc \theta ={(2-e^2)}^{3/2}$.

Answer 1

${\tan }^2\theta =1-e^2$ (given).
Now $\sec \theta +{\tan }^3\theta \csc \theta$
$=\sec \theta (1+{\tan }^3\theta \csc \theta /\sec \theta )$
$=\sec \theta (1+{\tan }^3\theta \cot \theta )={\sec }^3\theta$
$={(1+{\tan }^2\theta )}^{3/2}=(1+1-e^2{)}^{3/2}=(2-e^2{)}^{3/2}$