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Verify the following:

$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$

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Hint:$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$ $$\frac{2\cot x}{\tan 2x}=\frac{1-\tan ^2x}{\tan ^2x}=\frac{1}{\tan ^2x}-1=\frac{cos^2x}{sin^2x}-1=$$$$=\frac{1-sin^2x}{sin^2x}-1= \csc^2x-2$$

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Another way (convert to sines and cosines):

$$\frac{2 \cot x}{\tan 2x}=\frac{2\cos x}{\sin x} \frac{\cos 2x}{\sin 2x}$$

$$\cos 2x=\cos ^2x-\sin ^2x$$

$$\sin 2x=2\sin x \cos x$$

$$\csc ^2x=\frac{1}{\sin^2x}$$

From here, just put it together!


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Using the same technique as presented here (where it explains why we want to do things this way) we put $z = e^{ix}$ to obtain $$ \cot x = i \frac{z + \frac{1}{z}}{z - \frac{1}{z}} = i \frac{z^2+1}{z^2-1}$$ and $$ \tan 2x = \frac{1}{i} \frac{z^2 - \frac{1}{z^2}}{z^2 + \frac{1}{z^2}} = \frac{1}{i} \frac{z^4 - 1}{z^4 + 1}.$$ and $$ \csc^2 x - 2 = \left( \frac{2i}{z - \frac{1}{z} }\right)^2 -2 = -4 \left( \frac{1}{z - \frac{1}{z}}{} \right)^2 -2 = -4 \left(\frac{z}{z^2-1} \right)^2 - 2 \\ = \frac{-4z^2}{(z^2-1)^2} -2 = -2 \frac{z^4+1}{(z^2-1)^2}$$ But now $$ \frac{2\cot x}{\tan 2x} = \frac{ 2i \frac{z^2+1}{z^2-1}}{\frac{1}{i} \frac{z^4 - 1}{z^4 + 1}} = -2 \frac{(z^2+1)(z^4+1)}{(z^4-1)(z^2-1)} = -2 \frac{z^4+1}{(z^2-1)(z^2-1)}$$ and we are done.

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