Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)

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Show how ((cos^2(x)-sin^2(x))/sin(x))-2= sin^2(x)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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.

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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
no questions?
What do you suggest I do in order to get them equal?
they are not 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

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