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I'm having a hard time proving the following:

$$(\cot(x) - \csc(x))^2 = \frac{(\sec(x) - 1)}{(\sec(x) + 1)}$$

As far as I can tell this specific question has not been asked, but please let me know if this is a duplicate.

I am trying to manipulate the LHS to equal the RHS. I have tried some rearranging but I don't know if I'm getting anywhere This is what I have tried so far:

$$= \cot^2(x) - 2\cot(x)\csc(x) + \csc^2(x)$$

$$=\cot^2(x) - 2\cot(x)\csc(x) + 1 + \cot^2(x)$$

$$=\frac{1}{\tan^2(x)} - \frac{2\csc(x)}{\tan(x)} + 1 + \frac{1}{\tan^2(x)}$$

$$=\frac{2}{\tan^2(x)} - \frac{2\csc(x)\tan(x) + \tan^2(x)}{\tan^2(x)}$$

$$=\frac{2 - 2\csc(x)\tan(x) + \sec^2(x) - 1}{\sec^2(x) - 1}$$

$$=\frac{2 - 2\tan(x) + \sec^2(x) - 1}{\sec^2(x) - 1}$$

I feel a vague sense that I am getting closer, but I've already spent half an hour on this question. Is this the right approach, or is it completely circuitous? I am getting close? If anyone could point me in the right direction, that would be much appreciated.

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If you want to go on with your idea there is a mistake here (look the $\color{red}{\text{red color}}$):

$$\frac{1}{\tan^2(x)} - \frac{2\csc(x)}{\tan(x)} + 1 + \frac{1}{\tan^2(x)}=\frac{2}{\tan^2(x)} - \frac{2\csc(x)\tan(x) \color{red}{+} \tan^2(x)}{\tan^2(x)}$$

The right one is

$$\frac{1}{\tan^2(x)} - \frac{2\csc(x)}{\tan(x)} + 1 + \frac{1}{\tan^2(x)}=\frac{2}{\tan^2(x)} - \frac{2\csc(x)\tan(x) \color{red}{-} \tan^2(x)}{\tan^2(x)}$$

Now if you go on you will have:

$$\frac{2-2\csc(x)\tan(x) + \tan^2(x)}{\tan^2(x)}=\frac{2-2\sec(x)+ \sec^2(x)-1}{\sec^2(x)-1}=\frac{2(1-\sec(x))+ (\sec(x)-1)(\sec(x)+1)}{(\sec(x)-1)(\sec(x)+1)}=\frac{\sec x -1}{\sec x +1}$$

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  • $\begingroup$ Awesome. Thanks so much, silly typos are the worst :/ $\endgroup$ – Benjamin Lowry Jan 9 '17 at 13:23
  • $\begingroup$ you are very welcome! $\endgroup$ – Arnaldo Jan 9 '17 at 13:24
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To prove
${{\left( \cot x-\csc x \right)}^{2}}=\frac{\sec x-1}{\sec x+1}$
Start with R.H.S,
$R.H.S=\frac{\sec x-1}{\sec x+1}\times \frac{\sec x-1}{\sec x-1}$
Rationalization gives,
$\begin{align} & R.H.S=\frac{{{\left( \sec x-1 \right)}^{2}}}{{{\sec }^{2}}x-1}=\frac{{{\left( \sec x-1 \right)}^{2}}}{{{\tan }^{2}}x}={{\left( \frac{\sec x-1}{\tan x} \right)}^{2}}={{\left( \frac{\frac{1}{\cos x}}{\frac{\sin x}{\cos x}}-\frac{1}{\tan x} \right)}^{2}} \\ & ={{\left( \frac{1}{\sin x}-\cot x \right)}^{2}}={{\left( \csc x-\cot x \right)}^{2}}={{\left( \cot x-\csc x \right)}^{2}}=L.H.S \end{align}$
Proved.

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\begin{align} (\cot x - \csc x)^2&=\left(\frac{\cos x}{\sin x} - \frac{1}{\sin x}\right)^2\\ &= \frac{(\cos x -1)^2}{(\sin x)^2}\\ &= \frac{(\cos x -1)^2}{1-\cos ^2 x}\\ &= \frac{(1-\cos x )^2}{(1-\cos x)(1+\cos x)}\\ &= \frac{1-\cos x}{1+\cos x}\\ &= \frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}\\ &= \frac{\sec x -1}{\sec x +1} \end{align}

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  • $\begingroup$ Great answer. It's much more concise. Went with the other answer though as it showed me what I was doing wrong, but thanks a lot anyway! $\endgroup$ – Benjamin Lowry Jan 9 '17 at 13:24
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I'm going with this: $$\left(\cot(x)-\csc(x)\right)^2=\left(\frac{\cos(x)}{\sin(x)}-\frac{1}{\sin(x)}\right)^2= \frac{(\cos(x)-1)^2}{\sin^2(x)}$$ $$\frac{\sec(x)-1}{\sec(x)+1}=\frac{\frac{1}{\cos(x)}-1}{\frac{1}{\cos(x)}+1}=\frac{1-cos(x)}{1+\cos(x)}=\frac{(1-\cos(x))^2}{(1+\cos(x))(1-\cos(x))}=\frac{(\cos(x)-1)^2}{\sin^2(x)}.$$

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Use $\cos2y=2\cos^2y-1=1-2\sin^2y$ and $\sin2y=2\sin y\cos y$

$$\cot2y-\csc2y=\dfrac{\cos2y-1}{\sin2y}=\dfrac{-\sin^2y}{2\sin y\cos y}=-\tan y$$

$$\dfrac{\sec2y-1}{\sec2y+1}=\dfrac{1-\cos2y}{1+\cos2y}=\dfrac{2\sin^2y}{2\cos^2y}=?$$

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