Đáp án:

\(\begin{array}{l}
2.1\\
M = \dfrac{7}{2}\\
N =  - \dfrac{3}{4}\\
2.2\\
\sin \alpha  < \tan \alpha \\
\cos \alpha  < \cot \alpha \\
2.3\\
\left[ \begin{array}{l}
\sin \alpha  = \dfrac{{\sqrt {21} }}{5};\,\,\,\tan \alpha  = \dfrac{{\sqrt {21} }}{2};\,\,\,\cot \alpha  = \dfrac{{2\sqrt {21} }}{{21}}\\
\sin \alpha  =  - \dfrac{{\sqrt {21} }}{5};\,\,\,\tan \alpha  =  - \dfrac{{\sqrt {21} }}{2};\,\,\,\cot \alpha  =  - \dfrac{{2\sqrt {21} }}{{21}}
\end{array} \right.\\
2.4\\
\sin \alpha  = \dfrac{3}{5};\cos \alpha  = \dfrac{4}{5}\\
2.5\\
\tan B = \dfrac{5}{{12}}\\
2.6\\
a,\\
\dfrac{{\cos \alpha }}{{1 - \sin \alpha }} = \dfrac{{1 + \sin \alpha }}{{\cos \alpha }}\\
b,\\
\dfrac{{{{\left( {\sin \alpha  + \cos \alpha } \right)}^2} - {{\left( {\sin \alpha  - \cos \alpha } \right)}^2}}}{{\sin \alpha .\cos \alpha }} = 4
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2.1\\
\sin x = \cos \left( {90^\circ  - x} \right)\\
\cos x = \sin \left( {90^\circ  - x} \right)\\
{\sin ^2}x + {\cos ^2}x = 1\\
a,\\
M = {\cos ^2}15^\circ  + {\cos ^2}25^\circ  + {\cos ^2}35^\circ  + {\cos ^2}45^\circ  + {\cos ^2}55^\circ  + {\cos ^2}65^\circ  + {\cos ^2}75^\circ \\
 = {\cos ^2}15^\circ  + {\cos ^2}25^\circ  + {\cos ^2}35^\circ  + {\cos ^2}45^\circ  + {\sin ^2}\left( {90^\circ  - 55^\circ } \right) + {\sin ^2}\left( {90^\circ  - 65^\circ } \right) + {\sin ^2}\left( {90^\circ  - 75^\circ } \right)\\
 = {\cos ^2}15^\circ  + {\cos ^2}25^\circ  + {\cos ^2}35^\circ  + {\cos ^2}45^\circ  + {\sin ^2}35^\circ  + {\sin ^2}25^\circ  + {\sin ^2}15^\circ \\
 = \left( {{{\cos }^2}15^\circ  + {{\sin }^2}15^\circ } \right) + \left( {{{\cos }^2}25^\circ  + {{\sin }^2}25^\circ } \right) + \left( {{{\cos }^2}35^\circ  + {{\sin }^2}35^\circ } \right) + {\cos ^2}45^\circ \\
 = 1 + 1 + 1 + {\left( {\dfrac{{\sqrt 2 }}{2}} \right)^2}\\
 = 3 + \dfrac{1}{2}\\
 = \dfrac{7}{2}\\
N = {\sin ^2}10^\circ  - {\sin ^2}20^\circ  + {\sin ^2}30^\circ  - {\sin ^2}40^\circ  - {\sin ^2}50^\circ  - {\sin ^2}70^\circ  + {\sin ^2}80^\circ \\
 = {\sin ^2}10^\circ  - {\sin ^2}20^\circ  + {\sin ^2}30^\circ  - {\sin ^2}40^\circ  - {\cos ^2}\left( {90^\circ  - 50^\circ } \right) - {\cos ^2}\left( {90^\circ  - 70^\circ } \right) + {\cos ^2}\left( {90^\circ  - 80^\circ } \right)\\
 = {\sin ^2}10^\circ  - {\sin ^2}20^\circ  + {\sin ^2}30^\circ  - {\sin ^2}40^\circ  - {\cos ^2}40^\circ  - {\cos ^2}20^\circ  + {\cos ^2}10^\circ \\
 = \left( {{{\sin }^2}10^\circ  + {{\cos }^2}10^\circ } \right) - \left( {{{\sin }^2}20^\circ  + {{\cos }^2}20^\circ } \right) - \left( {{{\sin }^2}40^\circ  + {{\cos }^2}40^\circ } \right) + {\sin ^2}30^\circ \\
 = 1 - 1 - 1 + {\left( {\dfrac{1}{2}} \right)^2}\\
 =  - 1 + \dfrac{1}{4}\\
 =  - \dfrac{3}{4}\\
2.2\\
0^\circ  < \alpha  < 90^\circ  \Rightarrow \left\{ \begin{array}{l}
0 < \sin \alpha  < 1\\
0 < \cos \alpha  < 1
\end{array} \right.\\
*)\\
\sin \alpha  - \tan \alpha  = \sin \alpha  - \dfrac{{\sin \alpha }}{{\cos \alpha }} = \sin \alpha .\left( {1 - \dfrac{1}{{\cos \alpha }}} \right)\\
 = \sin \alpha .\dfrac{{\cos \alpha  - 1}}{{\cos \alpha }}\\
0 < \cos \alpha  < 1 \Rightarrow \cos \alpha  - 1 < 0\\
 \Rightarrow \sin \alpha .\dfrac{{\cos \alpha  - 1}}{{\cos \alpha }} < 0\\
 \Rightarrow \sin \alpha  - \tan \alpha  < 0\\
 \Leftrightarrow \sin \alpha  < \tan \alpha \\
*)\\
\cos \alpha  - \cot \alpha  = \cos \alpha  - \dfrac{{\cos \alpha }}{{\sin \alpha }} = \cos \alpha .\left( {1 - \dfrac{1}{{\sin \alpha }}} \right)\\
 = \cos \alpha .\dfrac{{\sin \alpha  - 1}}{{\sin \alpha }}\\
0 < \sin \alpha  < 1 \Rightarrow \sin \alpha  - 1 < 0\\
 \Rightarrow \cos \alpha .\dfrac{{\sin \alpha  - 1}}{{\sin \alpha }} < 0\\
 \Rightarrow \cos \alpha  - \cot \alpha  < 0\\
 \Leftrightarrow \cos \alpha  < \cot \alpha \\
2.3\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {0,4^2} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + {\left( {\dfrac{2}{5}} \right)^2} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  + \dfrac{4}{{25}} = 1\\
 \Leftrightarrow {\sin ^2}\alpha  = \dfrac{{21}}{{25}}\\
 \Rightarrow \left[ \begin{array}{l}
\sin \alpha  = \dfrac{{\sqrt {21} }}{5}\\
\sin \alpha  =  - \dfrac{{\sqrt {21} }}{5}
\end{array} \right.\\
*)\\
\sin \alpha  = \dfrac{{\sqrt {21} }}{5} \Rightarrow \left\{ \begin{array}{l}
\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{\sqrt {21} }}{2}\\
\cot \alpha  = \dfrac{1}{{\tan \alpha }} = \dfrac{2}{{\sqrt {21} }} = \dfrac{{2\sqrt {21} }}{{21}}
\end{array} \right.\\
*)\\
\sin \alpha  =  - \dfrac{{\sqrt {21} }}{5} \Rightarrow \left\{ \begin{array}{l}
\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} =  - \dfrac{{\sqrt {21} }}{2}\\
\cot \alpha  = \dfrac{1}{{\tan \alpha }} =  - \dfrac{2}{{\sqrt {21} }} =  - \dfrac{{2\sqrt {21} }}{{21}}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sin \alpha  = \dfrac{{\sqrt {21} }}{5};\,\,\,\tan \alpha  = \dfrac{{\sqrt {21} }}{2};\,\,\,\cot \alpha  = \dfrac{{2\sqrt {21} }}{{21}}\\
\sin \alpha  =  - \dfrac{{\sqrt {21} }}{5};\,\,\,\tan \alpha  =  - \dfrac{{\sqrt {21} }}{2};\,\,\,\cot \alpha  =  - \dfrac{{2\sqrt {21} }}{{21}}
\end{array} \right.\\
2.4\\
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 \alpha  = \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 a - \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};\cos \alpha  = \dfrac{4}{5}\\
2.5\\
Tam\,\,giác\,\,ABC\,\,vuông\,\,tại\,\,C\,\,nên:\,\,\widehat A + \widehat B = 90^\circ \\
 \Rightarrow \sin B = \cos \left( {90^\circ  - B} \right) = \cos A = \dfrac{5}{{13}}\\
0 < \widehat B < 90^\circ  \Rightarrow 0 < \cos B < 1\\
{\sin ^2}B + {\cos ^2}B = 1\\
 \Leftrightarrow {\left( {\dfrac{5}{{13}}} \right)^2} + {\cos ^2}B = 1\\
 \Leftrightarrow {\cos ^2}B = \dfrac{{144}}{{169}}\\
0 < \cos B < 1 \Rightarrow \cos B = \dfrac{{12}}{{13}}\\
 \Rightarrow \tan B = \dfrac{{\sin B}}{{\cos B}} = \dfrac{5}{{12}}\\
2.6\\
a,\\
\dfrac{{\cos \alpha }}{{1 - \sin \alpha }} = \dfrac{{\cos \alpha \left( {1 + \sin \alpha } \right)}}{{\left( {1 - \sin \alpha } \right)\left( {1 + \sin \alpha } \right)}}\\
 = \dfrac{{\cos \alpha \left( {1 + \sin \alpha } \right)}}{{1 - {{\sin }^2}\alpha }} = \dfrac{{\cos \alpha \left( {1 + \sin \alpha } \right)}}{{{{\cos }^2}\alpha }}\\
 = \dfrac{{1 + \sin \alpha }}{{\cos \alpha }}\\
b,\\
\dfrac{{{{\left( {\sin \alpha  + \cos \alpha } \right)}^2} - {{\left( {\sin \alpha  - \cos \alpha } \right)}^2}}}{{\sin \alpha .\cos \alpha }}\\
 = \dfrac{{\left( {{{\sin }^2}\alpha  + 2\sin \alpha .\cos \alpha  + {{\cos }^2}\alpha } \right) - \left( {{{\sin }^2}\alpha  - 2\sin \alpha .\cos \alpha  + {{\cos }^2}\alpha } \right)}}{{\sin \alpha .\cos \alpha }}\\
 = \dfrac{{\left( {1 + 2\sin \alpha .\cos \alpha } \right) - \left( {1 - 2\sin \alpha .\cos \alpha } \right)}}{{\sin \alpha .\cos \alpha }}\\
 = \dfrac{{1 + 2\sin \alpha .\cos \alpha  - 1 + 2\sin \alpha .\cos \alpha }}{{\sin \alpha .\cos \alpha }}\\
 = \dfrac{{4\sin \alpha .\cos \alpha }}{{\sin \alpha .\cos \alpha }}\\
 = 4
\end{array}\)