Question

If ${\tan }^2\theta =1-e^2$, then prove that $sec\theta +ta{n}^3\theta cosec\theta =(2-e^2{)}^{\frac{3}{2}}$

Answer 1

$\to \tan \theta =\sqrt{1-e^2}$

$\to \sec \theta +{\tan }^3\theta cosec\theta$

$\to \frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^3\theta }$

$\to {\sec }^3\theta$

$\to {\left(\sec \theta \right)}^3$

$\to {\left(\sqrt{1+{\tan }^2\theta }\right)}^3$

$\to {(1+1-e^2)}^{3/2}$

$\to {(2-e^2)}^{3/2}$