I keep coming to a dead end when working this one.
$\displaystyle csc x + cot x = \frac{1}{csc x - cot x}$
Thanks!
Work with the right hand side. Substitute csc x = 1/sin x and cot x = cos x/sin x. Simplify. Note that $\displaystyle \displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}$. If you need more help, please show - clearly - all your work and say where you get stuck.
So that gives me:
$\displaystyle \frac{\frac{sin(x)}{1-cos(x)}}{\frac{sin(x)}{sin(x)}} = \frac{sin^2(x)}{(1-cos(x)) (sin(x))}$
I'm not quite sure if that's what you meant by "divide everything". This doesn't seem to help.
Maybe...
$\displaystyle \frac{(1 - cos(x))(1 + cos(x))}{(1-cos(x)) (sin(x))} = \frac{1 + cos(x)}{sin(x)}$
Am I on the right track?
What part? Where you said to show my work and where I got stuck? I did show my work and where I got stuck...
I also see where you said:
I'm not sure how this is true or how I would apply it to the problem. Does it represent an identity?Note that $\displaystyle \displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}$
I see how Educated's tip solved the problem, I'm just not sure how I'd have figured that out on my own. It seems arbitrary to multiply everything by $\displaystyle \frac{sin(x)}{sin(x)}$ until you actually do it.
So if I were to receive this problem on a test, how would I know to do that? Identities are very frustrating for me. It often seems like trying random things until something works.
Just a note: in LaTeX when you use \frac it gives you smaller fractions eg. $\displaystyle \frac{\sin x}{\sin x}$. This gets really hard to read if you have fractions on fractions. Use \dfrac to give bigger fractions eg. $\displaystyle \dfrac{\sin x}{\sin x}$
What I meant was $\displaystyle \frac{\sin x}{1- \cos x} \div \frac{\sin x}{\sin x}$ which would give you: $\displaystyle \dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}$
This gives you the answer straight away.
I noticed how close you were to proving the identity and realised that the numerator must be a 1. So if you divided the numerator by sin(x) it will give you 1.
Strangely, when I work it this way, I don't get quite the same answer...
$\displaystyle \dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}=\dfrac{1}{\csc x - \cot x}$
That way doesn't seem to work, though I'm sure there's something I'm missing. I'd like to understand!
I don't think the missing step below should be too difficult to think of in this sort of topic:
$\displaystyle \displaystyle \frac{1}{1 - \cos (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} = ....$
Doing this is more suggestive (I would have thought) than the approach you ended up taking (which pretty much pulls a rabbit out of the hat). You do it as the next step from what you had way back in post #3.