Giải thích các bước giải:

\(\begin{array}{l}
1,\\
\tan a = 2 \Leftrightarrow \dfrac{{\sin a}}{{\cos a}} = 2 \Leftrightarrow \sin a = 2\cos a\\
{\sin ^2}a + {\cos ^2}a = 1\\
 \Leftrightarrow {\left( {2\cos a} \right)^2} + {\cos ^2}a = 1\\
 \Leftrightarrow {\cos ^2}a = \dfrac{1}{5}\\
\sin 2a = 2\sin a.\cos a = 2.\cos a.\cos a = 2.{\cos ^2}a = \dfrac{2}{5}\\
\cos 2a = 2{\cos ^2}a - 1 = 2.\dfrac{1}{5} - 1 =  - \dfrac{3}{5}\\
\tan 2a = \dfrac{{\sin 2a}}{{\cos 2a}} =  - \dfrac{2}{3}\\
2,\\
A = \cos 20^\circ .\cos 40^\circ .\cos 60^\circ .\cos 80^\circ \\
 \Leftrightarrow A.\sin 20^\circ  = \sin 20^\circ .cos20^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\
 \Leftrightarrow A.\sin 20^\circ  = \dfrac{1}{2}\sin 40^\circ .\cos 40^\circ .\dfrac{1}{2}.\cos 80^\circ \\
 \Leftrightarrow A.\sin 20^\circ  = \dfrac{1}{4}.\dfrac{1}{2}\sin 80^\circ .\cos 80^\circ \\
 \Leftrightarrow A.\sin 20^\circ  = \dfrac{1}{8}.\dfrac{1}{2}.sin160^\circ \\
 \Leftrightarrow A.\sin 20^\circ  = \dfrac{1}{{16}}.\sin \left( {180^\circ  - 160^\circ } \right)\\
 \Leftrightarrow A.\sin 20^\circ  = \dfrac{1}{{16}}.\sin 20^\circ \\
 \Leftrightarrow A = \dfrac{1}{{16}}
\end{array}\)