Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {\tan 2a - \tan a} \right)\left( {\sin 2a - \tan a} \right)\\
= \tan 2a.\sin 2a - \tan 2a.\tan a - \tan a.\sin 2a + {\tan ^2}a\\
= \dfrac{{{{\sin }^2}2a}}{{\cos 2a}} - \dfrac{{\sin 2a.\sin a}}{{\cos 2a.\cos a}} - \dfrac{{\sin a.\sin 2a}}{{\cos a}} + {\tan ^2}a\\
= \dfrac{{{{\sin }^2}2a.\cos a - \sin 2a.\sin a - \sin a.sin2a.\cos 2a}}{{\cos 2a.\cos a}} + {\tan ^2}a\\
= \dfrac{{{{\left( {2\sin a.\cos a} \right)}^2}.\cos a - 2\sin a.\cos a.\sin a - \sin a.2\sin a.\cos a.\cos 2a}}{{\cos 2a.\cos a}} + {\tan ^2}a\\
= \dfrac{{4{{\sin }^2}a.{{\cos }^3}a - 2{{\sin }^2}a.\cos a - 2{{\sin }^2}a.\cos a.\cos 2a}}{{\cos 2a.\cos a}} + {\tan ^2}a\\
= \dfrac{{2.{{\sin }^2}a.\cos a.\left( {2{{\cos }^2}a - 1 - \cos 2a} \right)}}{{\cos 2a.\cos a}} + {\tan ^2}a\\
= \dfrac{{2{{\sin }^2}a.\cos a.\left( {\cos 2a - \cos 2a} \right)}}{{\cos 2a.\cos a}} + {\tan ^2}a\\
= 0 + {\tan ^2}a = {\tan ^2}a
\end{array}\)
Em xem lại đề câu b nhé!!
