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