Here's the question you clicked on:
Romero
Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)
Yeah can someone show me the steps or what trig identities you used.
\[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\]
it's not you, latex is not rendering properly.
You know you can edit your question right?
I see the latex just fine
\[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)-2}=\sin^2(x) ?\]
Aha, hard refresh works!
I'm doing eigen values and eigen functions so the left expression has to equal the right.
You said you wrote it wrong? Did I write it right?
Sorry I deleted the wrong one already. The one I wrote is right. Anyways I should rephrase my questions. I want the left side to equal the right side. How can I change the left side in order to make the equation true so that the left hand side equals the right hand side?
\[\frac{1-\sin^2(x)-\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)}{\sin(x)}-2=\frac{1-2\sin^2(x)-2\sin(x)}{\sin(x)}\] hmmm.... I don't think they are the same .... Did you see if you could find a counterexample?
\[x=\frac{\pi}{2}\] \[\frac{\cos^2(x)-\sin^2(x)}{\sin(x)}-2=\sin^2(x)\] \[\frac{\cos^2(\frac{\pi}{2})-\sin^2(\frac{\pi}{2})}{\sin(\frac{\pi}{2})}=\frac{0-1}{1}=-1\] But the other side is 1 when x=pi/2 Therefore both sides are not the same
-1-2 does not equal 1
What do you suggest I do in order to get them equal?
this is not an identity I gave you a counterexample above try pluggin' in pi/2 for x we do not get the same thing on both sides unless you meant to type something else