Đáp án:
\(\begin{array}{l}
a,\,\,\,\,1\\
b,\,\,\,\,0\\
c,\,\,\,\,1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\sin ^6}a + {\cos ^6}a + 3{\sin ^2}a.{\cos ^2}a\\
= {\left( {{{\sin }^2}a} \right)^3} + {\left( {{{\cos }^2}a} \right)^3} + 3{\sin ^2}a.{\cos ^2}a\\
= \left( {{{\sin }^2}a + {{\cos }^2}a} \right).\left[ {{{\left( {{{\sin }^2}a} \right)}^2} - {{\sin }^2}a.{{\cos }^2}a + {{\left( {{{\cos }^2}a} \right)}^2}} \right] + 3{\sin ^2}a.{\cos ^2}a\\
= 1.\left[ {{{\left( {{{\sin }^2}a} \right)}^2} - {{\sin }^2}a.{{\cos }^2}a + {{\left( {{{\cos }^2}a} \right)}^2}} \right] + 3{\sin ^2}a.{\cos ^2}a\\
= {\left( {{{\sin }^2}a} \right)^2} - {\sin ^2}a.{\cos ^2}a + {\left( {{{\cos }^2}a} \right)^2} + 3{\sin ^2}a.{\cos ^2}a\\
= {\left( {{{\sin }^2}a} \right)^2} + 2.{\sin ^2}a.{\cos ^2}a + {\left( {{{\cos }^2}a} \right)^2}\\
= {\left( {{{\sin }^2}a + {{\cos }^2}a} \right)^2}\\
= {1^2}\\
= 1\\
b,\\
{\sin ^4}a - {\cos ^4}a - \left( {\sin a + \cos a} \right).\left( {\sin a - \cos a} \right)\\
= {\left( {{{\sin }^2}a} \right)^2} - {\left( {{{\cos }^2}a} \right)^2} - \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\\
= \left( {{{\sin }^2}a - {{\cos }^2}a} \right).\left( {{{\sin }^2}a + {{\cos }^2}a} \right) - \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\\
= \left( {{{\sin }^2}a - {{\cos }^2}a} \right).1 - \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\\
= \left( {{{\sin }^2}a - {{\cos }^2}a} \right) - \left( {{{\sin }^2}a - {{\cos }^2}a} \right)\\
= 0\\
c,\\
{\cos ^2}a + {\tan ^2}a.{\cos ^2}a\\
= {\cos ^2}a + {\left( {\dfrac{{\sin a}}{{\cos a}}} \right)^2}.{\cos ^2}a\\
= {\cos ^2}a + \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}.{\cos ^2}a\\
= {\cos ^2}a + {\sin ^2}a\\
= 1
\end{array}\)
