Prove following equation is an identity problem: $(1+ \cot x)^2 - 2\cot x = 1/((1-\cos x )(1+\cos x ))$

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?

• Yea, I messed up, it should be (1+ cot(x))^2 for the left side. – windy401 Mar 24 '15 at 18:41 2 Answers \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. • Wait, how do you get from line 3 to line 4 on the left side? – windy401 Mar 24 '15 at 20:06 • @windy401\sin^2x+\cos^2x=1\implies\sin^2x+\cos^2x-\cos^2x=1-\cos^2x\implies\sin^2x=1-{\cos}^2x$– 2012rcampion Mar 24 '15 at 21:12 You can't prove it because it's wrong. E.g., for$x=\pi/4$the left-hand side is zero while the right-hand side goes to infinity. • I messed up, it should be$(1+ cot(x))^2 for the left side. – windy401 Mar 24 '15 at 18:41
• @windy401: Still doesn't work out. – Matt L. Mar 24 '15 at 18:59
• 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. :/ – windy401 Mar 24 '15 at 19:06