Đáp án:
\(\begin{array}{l}
a,\,\,\,\,\dfrac{1}{3}\\
b,\,\,\,\dfrac{3}{2}\\
c,\,\,\,\,\dfrac{{81}}{{625}}\\
d,\,\,\,2\sin \alpha .\left( {1 + \cos \alpha } \right)\\
e,\,\,\,\,\dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha - \cos \alpha }}\\
f,\,\,\,{\cot ^2}\alpha
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\,\,\,\,\,{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\tan \alpha + \cot \alpha = 3\\
\Leftrightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\sin \alpha }} = 3\\
\Leftrightarrow \dfrac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = 3\\
\Leftrightarrow \dfrac{1}{{\sin \alpha .\cos \alpha }} = 3\\
\Leftrightarrow \sin \alpha .\cos \alpha = \dfrac{1}{3}\\
b,\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( {\dfrac{1}{2}} \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \dfrac{3}{4}\\
4{\cos ^2}\alpha - 6{\sin ^2}\alpha = 4.\dfrac{3}{4} - 6.{\left( {\dfrac{1}{2}} \right)^2} = 3 - \dfrac{3}{2} = \dfrac{3}{2}\\
c,\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\sin ^2}\alpha + {\left( {\dfrac{4}{5}} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}\alpha + \dfrac{{16}}{{25}} = 1\\
\Leftrightarrow {\sin ^2}\alpha = \dfrac{9}{{25}}\\
{\cos ^4}\alpha - {\cos ^2}\alpha + {\sin ^2}\alpha = {\left( {\dfrac{4}{5}} \right)^4} - {\left( {\dfrac{4}{5}} \right)^2} + \dfrac{9}{{25}} = \dfrac{{81}}{{625}}\\
d,\\
\dfrac{{2.{{\sin }^3}\alpha }}{{1 - \cos \alpha }} = \dfrac{{2\sin \alpha .{{\sin }^2}\alpha }}{{1 - \cos \alpha }} = \dfrac{{2.\sin \alpha \left( {1 - {{\cos }^2}\alpha } \right)}}{{1 - \cos \alpha }}\\
= \dfrac{{2\sin \alpha .\left( {1 - \cos \alpha } \right)\left( {1 + \cos \alpha } \right)}}{{1 - \cos \alpha }} = 2\sin \alpha .\left( {1 + \cos \alpha } \right)\\
e,\\
\dfrac{{1 + 2\sin \alpha .\cos \alpha }}{{{{\sin }^2}\alpha - {{\cos }^2}\alpha }} = \dfrac{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + 2\sin \alpha .\cos \alpha }}{{\left( {\sin \alpha - \cos \alpha } \right).\left( {\sin \alpha + \cos \alpha } \right)}}\\
= \dfrac{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2}}}{{\left( {\sin \alpha - \cos \alpha } \right).\left( {\sin \alpha + \cos \alpha } \right)}} = \dfrac{{\sin \alpha + \cos \alpha }}{{\sin \alpha - \cos \alpha }}\\
f,\\
\tan \alpha .\cot \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}.\dfrac{{\cos \alpha }}{{\sin \alpha }} = 1\\
\dfrac{{\tan \alpha }}{{\cot \alpha }} + \dfrac{{\cot \alpha }}{{\tan \alpha }} - \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\\
= \dfrac{{{{\tan }^2}\alpha + {{\cot }^2}\alpha }}{{\cot \alpha .\tan \alpha }} - {\left( {\dfrac{{\sin \alpha }}{{\cos \alpha }}} \right)^2}\\
= \dfrac{{{{\tan }^2}\alpha + {{\cot }^2}\alpha }}{1} - {\tan ^2}\alpha \\
= {\tan ^2}\alpha + {\cot ^2}\alpha - {\tan ^2}\alpha \\
= {\cot ^2}\alpha
\end{array}\)
