1. ## Proving an identity.

I keep coming to a dead end when working this one.

$csc x + cot x = \frac{1}{csc x - cot x}$

Thanks!

2. Originally Posted by JennyFlowers
I keep coming to a dead end when working this one.

$csc x + cot x = \frac{1}{csc x - cot x}$

Thanks!
Work with the right hand side. Substitute csc x = 1/sin x and cot x = cos x/sin x. Simplify. Note that $\displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}$. If you need more help, please show - clearly - all your work and say where you get stuck.

3. Thank you. Here's what I have so far:

$\frac{1}{csc x - cot x} = \frac{1}{\frac{1}{sin x}-\frac{cos x}{sin x}} = \frac{1}{\frac{1 - cos x}{sin x}}=\frac{sin x}{1-cos x}$

I'm not sure what to do at this point.

4. Divide everything by a cleverly disguised 1: $\, \dfrac{\sin (x)}{\sin (x)}$

5. So that gives me:

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

I'm not quite sure if that's what you meant by "divide everything". This doesn't seem to help.

Maybe...

$\frac{(1 - cos(x))(1 + cos(x))}{(1-cos(x)) (sin(x))} = \frac{1 + cos(x)}{sin(x)}$

Am I on the right track?

6. here is another way to approach it

$(\csc{x}+\cot{x})(\csc{x}-\cot{x}) = 1$

which is

$\csc^2{x}-\cot^2{x} =1$

Can you finish up from here?

7. Originally Posted by JennyFlowers
Thank you. Here's what I have so far:

$\frac{1}{csc x - cot x} = \frac{1}{\frac{1}{sin x}-\frac{cos x}{sin x}} = \frac{1}{\frac{1 - cos x}{sin x}}=\frac{sin x}{1-cos x}$

I'm not sure what to do at this point.
What you do is go back and read what I noted in my previous post.

8. Looks like I just about had it solved and didn't see it. As long as I'm right about Educated's tip:

$\frac{1 + cos(x)}{sin(x)} = csc x + cot x$

Did I do it right?

9. yep that is it

10. Originally Posted by mr fantastic
What you do is go back and read what I noted in my previous post.
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:

Note that $\displaystyle \frac{1}{1 - \cos (x)} = \frac{1 + \cos (x)}{\sin^2 (x)}$
I'm not sure how this is true or how I would apply it to the problem. Does it represent an identity?

11. 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 $\frac{sin(x)}{sin(x)}$ 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.

12. Just a note: in LaTeX when you use \frac it gives you smaller fractions eg. $\frac{\sin x}{\sin x}$. This gets really hard to read if you have fractions on fractions. Use \dfrac to give bigger fractions eg. $\dfrac{\sin x}{\sin x}$

Originally Posted by JennyFlowers
$\frac{\frac{sin(x)}{1-cos(x)}}{\frac{sin(x)}{sin(x)}} = \frac{sin^2(x)}{(1-cos(x)) (sin(x))}$

I'm not quite sure if that's what you meant by "divide everything".
What I meant was $\frac{\sin x}{1- \cos x} \div \frac{\sin x}{\sin x}$ which would give you: $\dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}$

This gives you the answer straight away.

Originally Posted by JennyFlowers
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 $\frac{sin(x)}{sin(x)}$ 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.
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.

13. Originally Posted by Educated
What I meant was $\frac{\sin x}{1- \cos x} \div \frac{\sin x}{\sin x}$ which would give you: $\dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}$
Strangely, when I work it this way, I don't get quite the same answer...

$\dfrac{\sin x \div \sin x }{1 \div \sin x - \cos x \div \sin x}=\dfrac{1}{\csc x - \cot x}$

That way doesn't seem to work, though I'm sure there's something I'm missing. I'd like to understand!

14. Originally Posted by JennyFlowers
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:

Note that

I'm not sure how this is true or how I would apply it to the problem. Does it represent an identity?
I don't think the missing step below should be too difficult to think of in this sort of topic:

$\displaystyle \frac{1}{1 - \cos (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} = ....$

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.

15. Originally Posted by mr fantastic
I don't think the missing step below should be too difficult to think of in this sort of topic:

$\displaystyle \frac{1}{1 - \cos (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} = ....$

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.
Sorry, what I meant was that the last step in post #3 was:

$\frac{\sin (x)}{1- \cos (x)}$

I'm not sure how that is the same as where you're starting from:

$\frac{1}{1 - \cos (x)}$

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