`cos4x.tan2x-sin4x` tính

$\begin{array}{l}
\cos 4x\tan 2x - \sin 4x\\
 = \dfrac{{\cos 4x\sin 2x}}{{\cos 2x}} - \sin 4x\\
 = \dfrac{{\cos 4x\sin 2x - \sin 4x\cos 2x}}{{\cos 2x}}\\
 = \dfrac{{\sin 2x\cos 4x - \sin 4x\cos 2x}}{{\cos 2x}}\\
 = \dfrac{{\sin \left( {2x - 4x} \right)}}{{\cos 2x}} = \dfrac{{\sin \left( { - 2x} \right)}}{{\cos 2x}} = \dfrac{{ - \sin 2x}}{{\cos 2x}}\\
 =  - \tan 2x
\end{array}$