If $\tan^2\theta = 1-e^2$,then the value of
$\sec\theta$ + $\tan^3\theta \cdot \csc\theta$ is…
NOTE:
$1/\cos\theta +\cos^3\theta/\sin^3\theta \cdot 1/\sin\theta$
multiply: $\sin^2 \theta/\cos \theta$ then we get
$\sin^2\theta/\cos^2\theta+\cos\theta \cdot 1/\tan\theta \cdot \tan\theta$
so $1-e^2 + \cos\theta$.
Best Answer
\begin{align*} \sec\theta+\tan^3\theta\csc\theta &=\frac1{\cos\theta}+\tan^2\theta\cdot\frac{\sin\theta}{\cos\theta}\cdot\frac1{\sin\theta} \\ &=\frac1{\cos\theta}\left(1+\tan^2\theta\right) \\ &=\sec\theta\sec^2\theta \end{align*}
Now $\sec^2\theta=1+\tan^2\theta=1+1-e^2$.