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I need to 'prove' that $$(1+ \cot(x))^2 - 2\cot(x) = \frac1{(1-\cos(x))(1+\cos(x))}$$

The book doesn't actually show answers for these types of problems, which hasn't been a problem till now, I've found the ones for far easy enough, but this one is stumping me. I know the $1 + \cot$ can be changed to csc, and got the right hand side down to $1/\sin^2$, but past that I'm stuck. Can someone point me in the right direction of what to do next? Or what I should have done if I'm way off with what I've done so far?

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    $\begingroup$ Yea, I messed up, it should be $(1+ cot(x))^2 for the left side. $\endgroup$ – windy401 Mar 24 '15 at 18:41
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$$ \begin{align} (1 + \cot x)^2 - 2\cot x &= \frac 1 {(1 - \cos x)(1 + \cos x)} \\ 1+2\cot x+\cot^2x-2\cot x&=\frac 1{1-\cos x+\cos x-\cos^2x} \\ 1 + \left(\frac{\cos x}{\sin x}\right)^2&=\frac 1{1-\cos^2x} \\ \frac{\sin^2 x+\cos^2x}{\sin^2x}&=\frac 1{\sin^2 x} \\ \frac{1}{\sin^2x}&=\frac 1{\sin^2 x} \end{align} $$ Q.E.D.

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  • $\begingroup$ Wait, how do you get from line 3 to line 4 on the left side? $\endgroup$ – windy401 Mar 24 '15 at 20:06
  • $\begingroup$ @windy401 $\sin^2x+\cos^2x=1$ $\implies\sin^2x+\cos^2x-\cos^2x=1-\cos^2x$ $\implies\sin^2x=1-{\cos}^2x$ $\endgroup$ – 2012rcampion Mar 24 '15 at 21:12
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You can't prove it because it's wrong. E.g., for $x=\pi/4$ the left-hand side is zero while the right-hand side goes to infinity.

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  • $\begingroup$ I messed up, it should be $(1+ cot(x))^2 for the left side. $\endgroup$ – windy401 Mar 24 '15 at 18:41
  • $\begingroup$ @windy401: Still doesn't work out. $\endgroup$ – Matt L. Mar 24 '15 at 18:59
  • $\begingroup$ Wow I am so sorry, it's cos on the right hand side, fixed in main post. Must have been more tired last night when I posted than I thought. :/ $\endgroup$ – windy401 Mar 24 '15 at 19:06

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