Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{1 - \sin 4x + 2\sin 3x.\cos x}}{{\sin x + \cos x - \sin x.\cos x - {{\sin }^2}x}}\\
= \dfrac{{1 - \sin 4x + \sin \left( {3x + x} \right) + \sin \left( {3x - x} \right)}}{{\left( {\sin x + \cos x} \right) - \sin x\left( {\cos x + \sin x} \right)}}\\
= \dfrac{{1 - \sin 4x + \sin 4x + \sin 2x}}{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x} \right)}}\\
= \dfrac{{1 + \sin 2x}}{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x} \right)}}\\
= \dfrac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x.\cos x}}{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x} \right)}}\\
= \dfrac{{{{\left( {\sin x + \cos x} \right)}^2}}}{{\left( {\sin x + \cos x} \right)\left( {1 - \sin x} \right)}}\\
= \dfrac{{\sin x + \cos x}}{{1 - \sin x}}
\end{array}\)
