Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{1 - \cos 4a}}{{\dfrac{1}{{{{\cos }^2}2a}} - 1}} + \dfrac{{1 + \cos 4a}}{{\dfrac{1}{{{{\sin }^2}2a}} - 1}}\\
= \dfrac{{1 - \left( {1 - 2{{\sin }^2}2a} \right)}}{{\dfrac{{1 - {{\cos }^2}2a}}{{{{\cos }^2}2a}}}} + \dfrac{{1 + \left( {2{{\cos }^2}2a - 1} \right)}}{{\dfrac{{1 - {{\sin }^2}2a}}{{{{\sin }^2}2a}}}}\\
= \dfrac{{2{{\sin }^2}2a}}{{\dfrac{{{{\sin }^2}2a}}{{{{\cos }^2}2a}}}} + \dfrac{{2{{\cos }^2}2a}}{{\dfrac{{{{\cos }^2}2a}}{{{{\sin }^2}2a}}}}\\
= \dfrac{{2{{\sin }^2}2a.{{\cos }^2}2a}}{{{{\sin }^2}2a}} + \dfrac{{2{{\cos }^2}2a.{{\sin }^2}2a}}{{{{\cos }^2}2a}}\\
= 2{\cos ^2}2a + 2{\sin ^2}2a\\
= 2\left( {{{\sin }^2}2a + {{\cos }^2}2a} \right)\\
= 2.1 = 2
\end{array}\)
