Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{1 + {{\sin }^4}a - {{\cos }^4}a}}{{1 + {{\sin }^6}a - {{\cos }^6}a}}\\
= \dfrac{{1 + \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\left( {{{\sin }^2}a + {{\cos }^2}a} \right)}}{{1 + \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\left( {{{\sin }^4}a + {{\sin }^2}a.{{\cos }^2}a + {{\cos }^4}a} \right)}}\\
= \dfrac{{1 + \left( {{{\sin }^2}a - {{\cos }^2}a} \right).1}}{{1 + \left( {{{\sin }^2}a - {{\cos }^2}a} \right).\left[ {{{\left( {{{\sin }^2}a + {{\cos }^2}a} \right)}^2} - {{\sin }^2}a.{{\cos }^2}a} \right]}}\\
= \dfrac{{1 + {{\sin }^2}a - {{\cos }^2}a}}{{1 + \left( {{{\sin }^2}a - {{\cos }^2}a} \right).\left( {{1^2} - {{\sin }^2}a.{{\cos }^2}a} \right)}}\\
= \dfrac{{{{\sin }^2}a + \left( {1 - {{\cos }^2}a} \right)}}{{1 + \left( {{{\sin }^2}a - \left( {1 - {{\sin }^2}a} \right)} \right).\left( {1 - {{\sin }^2}a\left( {1 - {{\sin }^2}a} \right)} \right)}}\\
= \dfrac{{{{\sin }^2}a + {{\sin }^2}a}}{{1 + \left( {2{{\sin }^2}a - 1} \right)\left( {1 - {{\sin }^2}a + {{\sin }^4}a} \right)}}\\
= \dfrac{{2{{\sin }^2}a}}{{1 + 2{{\sin }^2}a - 2{{\sin }^4}a + 2{{\sin }^6}a - 1 + {{\sin }^2}a - {{\sin }^4}a}}\\
= \dfrac{{2{{\sin }^2}a}}{{2{{\sin }^6}a - 3{{\sin }^4}a + 3{{\sin }^2}a}}\\
= \dfrac{2}{{2{{\sin }^4}a - 3{{\sin }^2}a + 3}}
\end{array}\)
