Đáp án:
$\begin{array}{l}
\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}}\\
= \frac{{2.\sin 2a.\cos 2a}}{{2.{{\cos }^2}2a}}.\frac{{\cos 2a}}{{2{{\cos }^2}a}}\\
= \frac{{\sin 2a}}{{2.{{\cos }^2}a}}\\
= \frac{{2.\sin a.\cos a}}{{2.{{\cos }^2}a}}\\
= \frac{{\sin a}}{{\cos a}}\\
= \tan a
\end{array}$
Vậy $\frac{{\sin 4a}}{{1 + \cos 4a}}.\frac{{\cos 2a}}{{1 + \cos 2a}} = \tan a$
