How do I manipulate this trig function in the correct way?

Question

Alright, so I've been at this question for a while and can't seem to get to the required answer. I don't know if there's some identity that I'm missing or if I'm just being dumb, but I've decided to try and get some help on here with it. Here's the question:

Show that

$$ \frac{\sec^2(x)-2}{\cos(x)+\sin(x)} = \sec(x)(\tan(x)-1) $$

I would really appreciate some help with this - thanks in advance!

Answer 1

\begin{align} \frac{\sec^2(x)-2}{\cos(x)+\sin(x)} &= \frac{1+\tan^2(x)-2}{\cos(x)+\sin(x)} \\[4pt] &= \frac{\tan^2(x)-1}{\cos(x)+\sin(x)} \\[4pt] &= \frac{(\tan(x)+1)(\tan(x)-1)}{\cos(x)+\sin(x)} \\[4pt] &= \frac{(\tan(x)+1)(\tan(x)-1)}{\cos(x)(1+\tan(x))} \\[4pt] &= \frac{\tan(x)-1}{\cos(x)} \\[4pt] &= \sec(x)(\tan(x)-1) \end{align}

Answer 2

The left-hand side of your equation is given by $$\frac{1-2\cos^2(x)}{\cos^2(x)(\sin(x)+\cos(x))}$$and the right-hand side: $$\frac{\sin(x)-\cos(x)}{\cos^2(x)}$$ Now use that $$(\sin(x)-\cos(x))(\sin(x)+\cos(x))=1-2\cos^2(x)$$