Question

If ${\tan }^2\theta =1-e^2$, then $sec\theta +{\tan }^3\theta \cos ec\theta$ is equal to

• A. $\sqrt{\left(2-e^2\right)}$
• B. ${\left(2-e^2\right)}^{\frac{3}{2}}$
• C. $\left(2-e^2\right)$
• D. None of these

Answer 1

The correct option is B

${\left(2-e^2\right)}^{\frac{3}{2}}$

Explanation for the correct option:

Step 1: Given that,

${\tan }^2\theta =1-e^2...\left(1\right)$

Then,

$\begin{array}{lcl}& \therefore & sec\theta +{\tan }^3\theta \cos ec\theta \\ & =& sec\theta +\left(\frac{{\sin }^3\theta }{{\cos }^3\theta }\right)\left(\frac{1}{\sin \theta }\right)\left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \right]\\ & =& sec\theta +\left({\tan }^2\theta sec\theta \right)\\ & =& sec\theta \left(1+{\tan }^2\theta \right)\\ & =& \sqrt{1+{\tan }^2\theta }\left(1+{\tan }^2\theta \right)\left[\because se{c}^2\theta =1+{\tan }^2\theta \right]\\ & =& {\left(1+{\tan }^2\theta \right)}^{\frac{3}{2}}\end{array}$

Step 2: From equation $\left(1\right)$,

$\begin{array}{rcl}& \therefore & {\left(1+1-e^2\right)}^{\frac{3}{2}}\\ & =& {\left(2-e^2\right)}^{\frac{3}{2}}\end{array}$

Therefore, the value of $sec\theta +{\tan }^3\theta \cos ec\theta =\begin{array}{rcl}& & \left(2-e^2\right)\end{array}\begin{array}{rcl}& & \end{array}$

Hence, the correct option is (B).