Verify that: $2\cot{x}/\tan{2}x = \csc^2x-2$ Verify the following: 
$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$
 A: Hint:$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\frac{2\cot x}{\tan 2x}=\frac{1-\tan ^2x}{\tan ^2x}=\frac{1}{\tan ^2x}-1=\frac{cos^2x}{sin^2x}-1=$$$$=\frac{1-sin^2x}{sin^2x}-1= \csc^2x-2$$
A: 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!
Edit:
You need this too:
$$\sin^2x+\cos^2x=1$$
A: 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.
