1) \(tan\alpha=\dfrac{2}{3}\)

Mà: \(tan\alpha\cdot cot\alpha=1\)

\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}\) 

Và: \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)

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

\(\Rightarrow cos\alpha=\sqrt{\dfrac{1}{1+tan^2\alpha}}=\sqrt{\dfrac{1}{1+\left(\dfrac{2}{3}\right)^2}}=\dfrac{3\sqrt{13}}{13}\)

Lại có:

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)

\(\Rightarrow sin\alpha=tan\alpha\cdot cos\alpha=\dfrac{2}{3}\cdot\dfrac{3\sqrt{13}}{13}=\dfrac{2\sqrt{13}}{13}\)

sin a=12/13

cos^2a=1-(12/13)^2=25/169

=>cosa=5/13

tan a=12/13:5/13=12/5

cot a=1:12/5=5/12

sin b=căn 3/2

cos^2b=1-(căn 3/2)^2=1/4

=>cos b=1/2

tan b=căn 3/2:1/2=căn 3

cot b=1/căn 3

có `cos α=1/2`

`=>cos^2 α=1/4`

Mà `cos^2 α +sin^2 α=1`

`=>1/4+sin^2 α=1`

`=>sin^2 α=1-1/4=3/4`

\(=>sin\alpha=\dfrac{\sqrt{3}}{2}\) (vì `sin α` >0)

ta có `sin α : cos α=tan α`

\(=>tan\alpha=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

ta có `tan α * cot α =1`

\(=>\sqrt{3}\cdot cot\alpha=1\\ =>cot\alpha=\dfrac{1}{\sqrt{3}}\)

tương tự ta có

\(\left\{{}\begin{matrix}sin\beta=\dfrac{\sqrt{2}}{2}\\cos\beta=1\\cot\beta=1\end{matrix}\right.\)

Bài 1:

a) Ta có:

\(tanB=\dfrac{AC}{AB}\Rightarrow\dfrac{AC}{AB}=\dfrac{5}{2}\)

\(\Rightarrow AC=\dfrac{AB\cdot5}{2}=\dfrac{6\cdot5}{2}=15\)  

b) Áp dụng Py-ta-go ta có: 

\(BC^2=AB^2+AC^2=6^2+15^2=261\)

\(\Rightarrow BC=\sqrt{261}=3\sqrt{29}\)

Bài 2: 

\(\left\{{}\begin{matrix}sinM=sin40^o\approx0,64\Rightarrow cosN\approx0,64\\cosM=cos40^o\approx0,77\Rightarrow sinN\approx0,77\\tanM=tan40^o\approx0,84\Rightarrow cotN\approx0,84\\cotM=cot40^o\approx1,19\Rightarrow tanN\approx1,19\end{matrix}\right.\)

a) \(\cos \alpha  =  - 0,75\)

\( \Leftrightarrow \alpha  ={138^ \circ }35'36''\) hay \(\alpha  =2,4188584\) rad

b) \(\tan \alpha  =  2,46\)

\( \Leftrightarrow \alpha  ={67^ \circ }52'01''\) hay \(\alpha  =1,1846956\) rad

c) \(\cot \alpha  =  -6,18\)

\( \Leftrightarrow \alpha  ={ -9^ \circ }11'30''\) hay \(\alpha  = -0,1604\) rad

a) sin a=0,8

Ta có: \(\sin^2a+\cos^2a=1\)

\(\Rightarrow\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\orbr{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

\(\sin a=0,8\)

\(\sin^2a=1-\sin^2a=1\)

\(\cos^2a=1-\sin^2a=1-0,8^2=0,36\)

\(\Rightarrow\hept{\begin{cases}\cos a=0,6\\\cos a=-0,6\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\tan a=\frac{4}{3}\\\tan a=\frac{-4}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\cot a=\frac{3}{4}\\\cot a=\frac{-3}{4}\end{cases}}\)

Code : Breacker

Vì \(\pi  < \alpha  < \frac{{3\pi }}{2}\)nên \(\sin \alpha  > 0\). Mặc khác, từ  \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\) suy ra

\(\sin \alpha  = \sqrt {1 - {{\cos }^2}\alpha }  = \sqrt {1 - \frac{4}{9}}  = \frac{{\sqrt 5 }}{3}\)

Do đó    \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{\sqrt 5 }}{3}}}{{ - \frac{2}{3}}} =  - \frac{{\sqrt 5 }}{2};\cot \alpha  = \frac{1}{{\tan \alpha }} = \frac{{ - 2}}{{\sqrt 5 }}\)

\(\tan\alpha=\dfrac{1}{3}\Leftrightarrow1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}\\ \Leftrightarrow\cos\alpha=\dfrac{1}{\sqrt{1+\tan^2\alpha}}=\dfrac{1}{\sqrt{\dfrac{9}{8}}}=\dfrac{2\sqrt{2}}{3}\\ \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\\ \tan\alpha=\dfrac{1}{3}\Leftrightarrow\cot\alpha=3\)