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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?

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Multiply numerator and denominator by $ (1+ \cos \theta) $ and simplify.

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$1= cos(\theta)^2 + sin(\theta)^2$ ~Pythagoras enter image description here

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