a) 

\(1+tan^2a=\frac{1}{cos^2a}\)           

\(1+3^2=\frac{1}{cos^2a}\)  

\(10=\frac{1}{cos^2a}\)     

\(cos^2a=\frac{1}{10}\) 

\(cosa=\pm\sqrt{\frac{1}{10}}=\pm\frac{1}{\sqrt{10}}\)    

\(sin^2a+cos^2a=1\)                                                             

\(sin^2a+\frac{1}{10}=1\)   

\(sin^2a=\frac{9}{10}\)     

\(sina=\pm\sqrt{\frac{9}{10}}=\pm\frac{3}{\sqrt{10}}\)   

Vì tan = 3 nên M có 2 trường hợp : 

TH1 : 

sin và cos cùng dương 

\(\Rightarrow M=\frac{\frac{1}{\sqrt{10}}+\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}}\)     

\(=\frac{\frac{4}{\sqrt{10}}}{-\frac{2}{\sqrt{10}}}\)                        

= -2 

TH2 : 

Cả sin và cos cùng âm 

\(\Rightarrow M=\frac{-\frac{1}{\sqrt{10}}+\left(-\frac{3}{\sqrt{10}}\right)}{-\frac{1}{\sqrt{10}}-\left(-\frac{3}{\sqrt{10}}\right)}\)            

=\(\frac{-\frac{4}{\sqrt{10}}}{\frac{2}{\sqrt{10}}}\)                 

= -2 

b) 

\(B=\frac{sin15+cos15}{cos15}-cot75\)         

=\(\frac{sin15}{cos15}+\frac{cos15}{cos15}-cot75\)          

=\(tan15+1-cot75\)     

=\(cot75+1-cot75\)    

= 1 

\(P=\dfrac{2sin\alpha-3cos\alpha}{3sin\alpha+2cos\alpha}\\ =\dfrac{\dfrac{2sin\alpha}{cos\alpha}-\dfrac{3cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}\\ =\dfrac{2tan\alpha-3}{3tan\alpha+2}=\dfrac{2.3-3}{3.3+2}=\dfrac{3}{11}\)

Ta có: \(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\quad (\alpha  \ne {90^o})\)

\( \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = 1 + {3^2} = 10\)

\( \Leftrightarrow {\cos ^2}\alpha  = \frac{1}{{10}} \Leftrightarrow \cos \alpha  =  \pm \frac{{\sqrt {10} }}{{10}}\)

Vì \({0^o} < \alpha  < {180^o}\) nên \(\sin \alpha  > 0\).

Mà \(\tan \alpha  = 3 > 0 \Rightarrow \cos \alpha  > 0 \Rightarrow \cos \alpha  = \frac{{\sqrt {10} }}{{10}}\)

Lại có: \(\sin \alpha  = \cos \alpha .\tan \alpha  = \frac{{\sqrt {10} }}{{10}}.3 = \frac{{3\sqrt {10} }}{{10}}.\)

\( \Rightarrow P = \dfrac{{2.\frac{{3\sqrt {10} }}{{10}} - 3.\frac{{\sqrt {10} }}{{10}}}}{{3.\frac{{3\sqrt {10} }}{{10}} + 2.\frac{{\sqrt {10} }}{{10}}}} = \dfrac{{\frac{{\sqrt {10} }}{{10}}\left( {2.3 - 3} \right)}}{{\frac{{\sqrt {10} }}{{10}}\left( {3.3 + 2} \right)}} = \dfrac{3}{{11}}.\)

Tham khảo:

a) 

Gọi M(x;y) là điểm trên đường tròn đơn vị sao cho \(\widehat {xOM} = \alpha \). Gọi N, P tương ứng là hình chiếu vuông góc của M lên các trục Ox, Oy.

Ta có: \(\left\{ \begin{array}{l}x = \cos \alpha \\y = \sin \alpha \end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}\alpha  = {x^2}\\{\sin ^2}\alpha  = {y^2}\end{array} \right.\)(1)

Mà \(\left\{ \begin{array}{l}\left| x \right| = ON\\\left| y \right| = OP = MN\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{x^2} = {\left| x \right|^2} = O{N^2}\\{y^2} = {\left| y \right|^2} = M{N^2}\end{array} \right.\)(2)

Từ (1) và (2) suy ra \({\sin ^2}\alpha  + {\cos ^2}\alpha  = O{N^2} + M{N^2} = O{M^2}\) (do \(\Delta OMN\) vuông tại N)

\( \Rightarrow {\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) (vì OM =1). (đpcm)

b) 

Ta có:  \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }}\;\;(\alpha  \ne {90^o})\)

\( \Rightarrow 1 + {\tan ^2}\alpha  = 1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}\)

Mà theo ý a) ta có \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) với mọi góc \(\alpha \)

\( \Rightarrow 1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\) (đpcm)

c) 

Ta có:  \(\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }}\;\;\;({0^o} < \alpha  < {180^o})\)

\( \Rightarrow 1 + {\cot ^2}\alpha  = 1 + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }}\)

Mà theo ý a) ta có \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) với mọi góc \(\alpha \)

\( \Rightarrow 1 + {\cot ^2}\alpha  = \frac{1}{{{{\sin }^2}\alpha }}\) (đpcm)

oh , bác sĩ ơi tui sắp chết

1.

\(\frac{1-2sin\alpha cos\alpha}{sin^2\alpha-cos^2\alpha}=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)

\(\Leftrightarrow\frac{1-2sin\alpha cos\alpha}{\left(sin\alpha-cos\alpha\right)\left(sin\alpha+cos\alpha\right)}=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)

\(\Leftrightarrow1-2sin\alpha cos\alpha=\left(sin\alpha-cos\alpha\right)^2\)

\(\Leftrightarrow1-2sin\alpha cos\alpha=sin^2\alpha+cos^2\alpha-2sin\alpha cos\alpha\)

\(\Leftrightarrow1-2sin\alpha cos\alpha=1-2sin\alpha cos\alpha\left(đpcm\right)\)

\(M=\frac{\frac{sina}{cosa}+\frac{cosa}{cosa}}{\frac{sina}{cosa}-\frac{cosa}{cosa}}=\frac{tana+1}{tana-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=...\)

\(N=\frac{\frac{sina.cosa}{cos^2a}}{\frac{sin^2a}{cos^2a}-\frac{cos^2a}{cos^2a}}=\frac{tana}{tan^2a-1}=...\) (thay số bấm máy)

\(P=\frac{\frac{sin^3a}{cos^3a}+\frac{cos^3a}{cos^3a}}{\frac{2sina.cos^2a}{cos^3a}+\frac{cosa.sin^2a}{cos^3a}}=\frac{tan^3a+1}{2tana+tan^2a}=...\)

a) Ta có:  \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) =  - \cos {16^o}\end{array} \right.\)

\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)

b) 

Ta có:

\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) =  - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) =  - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) =  - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha  < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha  - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha  - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha  - \cos \alpha .\tan \alpha .\cot \alpha \)

\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha  - \cos \alpha .1 = \cos \alpha .\)