Chứng mình rằng cotx - tanx - 2tan2x = 4cot4x

Giải thích các bước giải:

 Ta có:

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