## $\sin \theta + \frac{1}{\cos} \theta$

Jul 12, 2018

Let

$E \left(\theta\right) = \sin \theta + \frac{1}{\cos} \theta$

$E \left(\theta\right) = \frac{\sin \theta \cos \theta + 1}{\cos} \theta$

This form is already in terms of $\sin \theta$ and $\cos \theta$, but let's continue anyway.

Recognizing that $1$ is nothing more than ${\sin}^{2} \theta + {\cos}^{2} \theta$, we have

$E \left(\theta\right) = \frac{{\sin}^{2} \theta + \sin \theta \cos \theta + {\cos}^{2} \theta}{\cos} \theta$

We know that

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right) \implies {a}^{2} + a b + {b}^{2} = \frac{{a}^{3} - {b}^{3}}{a - b}$

$E \left(\theta\right) = \frac{\frac{{\sin}^{3} \theta - {\cos}^{3} \theta}{\sin \theta - \cos \theta}}{\cos} \theta = \frac{{\sin}^{3} \theta - {\cos}^{3} \theta}{\sin \theta \cos \theta - {\cos}^{2} \theta}$