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