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