<s>[INST] Proving $2\cot⁡4x = \cot⁡2x - \tan⁡2x$ How to prove the trigonometry equation is an identity?
$$2\cot⁡4x = \cot⁡2x - \tan⁡2x$$
Thank you in advance.
Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer. Thank you.
$2cot4x = cot2x - tan2x$
$2\frac{1}{tan 4x} = cot2x - tan2x$
$2\frac{1}{\frac{2 tan 2x}{1 - tan^2 2x}} = cot2x - tan2x$
$2\frac{1 - tan^2 2x}{2 tan 2x} = cot2x - tan2x$
$\frac{1}{tan 2x} -  \frac{tan^2 2x}{tan 2x} = cot2x - tan2x$
$cot2x - tan2x = cot2x - tan2x$
 [/INST] Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer.
$$\begin{align}
2\cot4x = \cot2x - \tan2x \\ 
2\cdot\frac{1}{\tan 4x} = \cot2x - \tan2x\\
2\cdot\frac{1}{\frac{2 \tan 2x}{1 - \tan^2 2x}} = \cot2x - \tan2x\\
2\cdot\frac{1 - \tan^2 2x}{2 \tan 2x} = \cot2x - \tan2x\\
\frac{1}{\tan 2x} -  \frac{\tan^2 2x}{tan 2x} = \cot2x - \tan2x\\
\cot2x - \tan2x = \cot2x - \tan2x\end{align}$$
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