giúp mình làm câu a 😢

$\begin{array}{l}
CMR:\cot x - \tan x = 2\cot 2x\\
\cot x - \tan x = \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} = \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x.\cos x}} = \dfrac{{\cos 2x}}{{\dfrac{1}{2}\sin 2x}} = 2\cot 2x\\
\cot x - \tan x - 2\tan 2x - 4\tan 4x\\
 = \left( {\cot x - \tan x} \right) - 2\tan 2x - 4\tan 4x\\
 = 2\cot 2x - 2\tan 2x - 4\tan 4x\\
 = 2\left( {2\cot 4x} \right) - 4\tan 4x\\
 = 4\cot 4x - 4\tan 4x\\
 = 4\left( {2.\cot 8x} \right) = 8\cot 8x
\end{array}$