Đáp án:

\(\begin{array}{l}
3,\\
a,\\
\cos \alpha  = 0,6\\
\tan \alpha  = \dfrac{4}{3}\\
\cot \alpha  = \dfrac{3}{4}\\
b,\\
\sin \alpha  = 0,8\\
\tan \alpha  = \dfrac{4}{3}\\
\cot \alpha  = \dfrac{3}{4}\\
c,\\
\cot \alpha  = \dfrac{1}{3}\\
\cos \alpha  = \dfrac{{\sqrt {10} }}{{10}}\\
\sin \alpha  = \dfrac{{3\sqrt {10} }}{{10}}\\
d,\\
\tan \alpha  = \dfrac{1}{2}\\
\sin \alpha  = \dfrac{{\sqrt 5 }}{5}\\
\cos \alpha  = \dfrac{{2\sqrt 5 }}{5}\\
4,\\
\cot \alpha  = \dfrac{4}{3}\\
5,\\
\tan B = \dfrac{5}{{12}}\\
6,\\
a,\,\,\,\,{\sin ^2}\alpha \\
b,\,\,\,2\\
c,\,\,\,{\sin ^3}\alpha \\
d,\,\,\,\,1\\
e,\,\,\,\,{\sin ^2}\alpha \\
f,\,\,\,\,1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
0^\circ  < \alpha  < 90^\circ  \Rightarrow \left\{ \begin{array}{l}
0 < \sin \alpha  < 1\\
0 < \cos \alpha  < 1
\end{array} \right.\\
a,\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {0,8^2} + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow 0,64 + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\cos ^2}\alpha  = 0,36\\
0 < \cos \alpha  < 1 \Rightarrow \cos \alpha  = 0,6\\
 \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{0,8}}{{0,6}} = \dfrac{4}{3}\\
\cot \alpha  = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{0,6}}{{0,8}} = \dfrac{3}{4}\\
b,\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {0,6^2} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + 0,36 = 1\\
 \Leftrightarrow {\sin ^2}\alpha  = 0,64\\
0 < \sin \alpha  < 1 \Rightarrow \sin \alpha  = 0,8\\
\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{0,8}}{{0,6}} = \dfrac{4}{3}\\
\cot \alpha  = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{{0,6}}{{0,8}} = \dfrac{3}{4}\\
c,\\
\tan \alpha  = 3 \Rightarrow \left\{ \begin{array}{l}
\cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{1}{3}\\
\dfrac{{\sin \alpha }}{{\cos \alpha }} = 3 \Leftrightarrow \sin \alpha  = 3\cos \alpha 
\end{array} \right.\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\left( {3\cos \alpha } \right)^2} + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow 9{\cos ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow 10{\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\cos ^2}\alpha  = \dfrac{1}{{10}}\\
0 < \cos \alpha  < 1 \Rightarrow \cos \alpha  = \dfrac{{\sqrt {10} }}{{10}}\\
\sin \alpha  = 3\cos \alpha  = \dfrac{{3\sqrt {10} }}{{10}}\\
d,\\
\cot \alpha  = 2 \Rightarrow \left\{ \begin{array}{l}
\tan \alpha  = \dfrac{1}{{\cot \alpha }} = \dfrac{1}{2}\\
\dfrac{{\cos \alpha }}{{\sin \alpha }} = 2 \Leftrightarrow \cos \alpha  = 2\sin \alpha 
\end{array} \right.\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {\left( {2\sin \alpha } \right)^2} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + 4{\sin ^2}\alpha  = 1\\
 \Leftrightarrow 5{\sin ^2}\alpha  = 1\\
 \Leftrightarrow {\sin ^2}\alpha  = \dfrac{1}{5}\\
0 < \sin \alpha  < 1 \Rightarrow \sin \alpha  = \dfrac{{\sqrt 5 }}{5}\\
\cos \alpha  = 2\sin \alpha  = \dfrac{{2\sqrt 5 }}{5}\\
4,\\
0 < \alpha  < 90^\circ  \Rightarrow \left\{ \begin{array}{l}
0 < \sin \alpha  < 1\\
0 < \cos \alpha  < 1
\end{array} \right.\\
\cos \alpha  - \sin \alpha  = \dfrac{1}{5}\\
 \Leftrightarrow \cos  = \sin \alpha  + \dfrac{1}{5}\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {\left( {\sin \alpha  + \dfrac{1}{5}} \right)^2} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {\sin ^2}\alpha  + \dfrac{2}{5}\sin \alpha  + \dfrac{1}{{25}} = 1\\
 \Leftrightarrow 2{\sin ^2}\alpha  + \dfrac{2}{5}\sin \alpha  - \dfrac{{24}}{{25}} = 0\\
 \Leftrightarrow {\sin ^2}\alpha  + \dfrac{1}{5}\sin \alpha  - \dfrac{{12}}{{25}} = 0\\
 \Leftrightarrow \left( {{{\sin }^2}\alpha  - \dfrac{3}{5}\sin \alpha } \right) + \left( {\dfrac{4}{5}\sin \alpha  - \dfrac{{12}}{{25}}} \right) = 0\\
 \Leftrightarrow \sin \alpha \left( {\sin \alpha  - \dfrac{3}{5}} \right) + \dfrac{4}{5}\left( {\sin \alpha  - \dfrac{3}{5}} \right) = 0\\
 \Leftrightarrow \left( {\sin \alpha  - \dfrac{3}{5}} \right)\left( {\sin \alpha  + \dfrac{4}{5}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \alpha  - \dfrac{3}{5} = 0\\
\sin \alpha  + \dfrac{4}{5} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin \alpha  = \dfrac{3}{5}\\
\sin \alpha  =  - \dfrac{4}{5}
\end{array} \right.\\
0 < \sin \alpha  < 1 \Rightarrow \sin \alpha  = \dfrac{3}{5} \Rightarrow \cos \alpha  = \dfrac{4}{5}\\
\cot \alpha  = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \dfrac{4}{3}\\
5,\\
Tam\,\,giác\,\,ABC\,\,vuông\,\,tại\,\,C\,\,nên:\\
\widehat A + \widehat B = 90^\circ  \Rightarrow \left\{ \begin{array}{l}
\sin B = \cos \left( {90^\circ  - B} \right) = \cos A\\
0 < \widehat B < 90^\circ  \Rightarrow 0 < \cos B < 1
\end{array} \right.\\
\sin B = \cos A = \dfrac{5}{{13}}\\
{\sin ^2}B + {\cos ^2}B = 1\\
 \Leftrightarrow {\left( {\dfrac{5}{{13}}} \right)^2} + {\cos ^2}B = 1\\
 \Leftrightarrow \dfrac{{25}}{{169}} + {\cos ^2}B = 1\\
 \Leftrightarrow {\cos ^2}B = 1 - \dfrac{{25}}{{169}}\\
 \Leftrightarrow {\cos ^2}B = \dfrac{{144}}{{169}}\\
0 < \cos B < 1 \Rightarrow \cos B = \dfrac{{12}}{{13}}\\
\tan B = \dfrac{{\sin B}}{{\cos B}} = \dfrac{5}{{12}}\\
6,\\
a,\\
\left( {1 - \cos \alpha } \right)\left( {1 + \cos \alpha } \right) = 1 - {\cos ^2}\alpha  = {\sin ^2}\alpha \\
b,\\
1 + {\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 + \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right) = 1 + 1 = 2\\
c,\\
\sin \alpha  - \sin \alpha .{\cos ^2}\alpha  = \sin \alpha .\left( {1 - {{\cos }^2}\alpha } \right) = \sin \alpha .{\sin ^2}\alpha  = {\sin ^3}\alpha \\
d,\\
{\sin ^4}\alpha  + {\cos ^4}\alpha  + 2{\sin ^2}\alpha .{\cos ^2}\alpha \\
 = \left( {{{\sin }^2}\alpha } \right) + 2.{\sin ^2}\alpha .{\cos ^2}\alpha  + {\left( {{{\cos }^2}\alpha } \right)^2}\\
 = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^2}\\
 = {1^2} = 1\\
e,\\
{\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}.{\cos ^2}\alpha \\
 = \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}.{\cos ^2}\alpha \\
 = {\sin ^2}\alpha \\
f,\\
{\cos ^2}\alpha  + {\tan ^2}\alpha .{\cos ^2}\alpha \\
 = {\cos ^2}\alpha .\left( {1 + {{\tan }^2}\alpha } \right)\\
 = {\cos ^2}\alpha .\left( {1 + \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}} \right)\\
 = {\cos ^2}\alpha .\dfrac{{{{\cos }^2}\alpha  + {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}\\
 = {\cos ^2}\alpha .\dfrac{1}{{{{\cos }^2}\alpha }}\\
 = 1
\end{array}\)