# Trigonometry identity problems

Ok, so I just learned trig identities and I come across this problem that had it's answer to it, and I have no idea how they got to that answer.

Here is the problem:

$$\frac{-\sec\theta}{1-\cos\theta}=\frac{-1-\sec\theta}{\sin^2\theta}$$

Now the problem calls for the left side to be adjusted. Here's where it came to first:

$$\frac{-1-\sec\theta}{1-\cos^2\theta}$$

After that, it then came to the solution which was:

$$\frac{-1-\sec\theta}{\sin^2\theta}$$

I'm stumped on how they got to the second step. What am I missing?

Multiply numerator and denominator by $(1+ \cos \theta)$ and simplify.
$1= cos(\theta)^2 + sin(\theta)^2$ ~Pythagoras