# Question #e7dd8

Jun 22, 2017

$\sec \theta \cdot \csc \theta$

#### Explanation:

You're right in thinking that it simplifies to $\tan \theta + \cot \theta$, but since that's not an option you have to look for alternate forms (other things it could be).

The first thing that comes to mind for me is to cleverly multiply by $1$ so we can make one single fraction.
$\sin \frac{\theta}{\cos} \theta \left(\sin \frac{\theta}{\sin} \theta\right) + \cos \frac{\theta}{\sin} \theta \left(\cos \frac{\theta}{\cos} \theta\right)$
which simplifies to
${\sin}^{2} \frac{\theta}{\cos \theta \cdot \sin \theta} + {\cos}^{2} \frac{\theta}{\cos \theta \cdot \sin \theta}$
which simplifies to
$\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos \theta \cdot \sin \theta}$

Now knowing the Pythagorean Identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$, this is equivalent to
$\frac{1}{\cos \theta \cdot \sin \theta}$
which is equivalent to
$\frac{1}{\cos} \theta \cdot \frac{1}{\sin} \theta$

Since $\sec \theta = \frac{1}{\cos} \theta$ and $\csc \theta = \frac{1}{\sin} \theta$, this equals
$\sec \theta \cdot \csc \theta$