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