Answer
$\sec(\theta)-\cos(-\theta)=\frac{\sin^{2}(\theta)}{\cos(\theta)}$
Work Step by Step
$\sec(\theta)$ can be represented as $\frac{1}{\cos(\theta)}$
Since cosine is an even function, $\cos(-\theta)=\cos(\theta)$
The equation can then be written as:
$\sec(\theta)-\cos(-\theta)=\frac{1}{\cos(\theta)}-\cos(\theta)$
The denominators of both terms can be made into $\cos(\theta)$ by multiplying $\cos(\theta)$ with $\frac{\cos(\theta)}{\cos(\theta)}$ to give $\frac{\cos^{2}(\theta)}{\cos(\theta)}$
$\frac{1}{\cos(\theta)}-\cos(\theta)=\frac{1-\cos^{2}(\theta)}{\cos(\theta)}$
BY using the identity $1-\cos^{2}(\theta)=\sin^{2}(\theta)$,
$\frac{1-\cos^{2}(\theta)}{\cos(\theta)}=\frac{\sin^{2}(\theta)}{\cos(\theta)}$
Therefore,
$\sec(\theta)-\cos(-\theta)=\frac{\sin^{2}(\theta)}{\cos(\theta)}$
