Thread: Woah, ok now, one more trig identity and I'm straight

1. Woah, ok now, one more trig identity and I'm straight

Alright, here's the problem:
sinxcosx + cos²x = cosx(1+cotx)/cscx

Ok, I can change some of this to end up w/the structure and same terms needed at the end, but I don't know how to end up exactly right. I only end up with-
cosx + 1+cotx[all over]/cscx

how do I get the cosx and 1+cotx to be a product of each other rather than a sum? Am I missing something obvious, or do I need to switch some things up and start out differently? It seems easy and like it can be done in just a couple or few steps, but idk if it's easier than I'm making it out to be and I'm missing something or if it's more than what I'm making it to be?

2. Hello, a1b2c3!

$\sin x\cos x + \cos^2\!x \:=\:\frac{\cos x(1+\cot x)}{\csc x}$

The right side is: . $\frac{1}{\csc x}\cdot\cos x\cdot(1 + \cot x)$

. . . . . . . . . . . $= \;\sin x\cdot\cos x\left(1 + \frac{\cos x}{\sin x}\right)$

Now multiply . . .