Cho biết \(\tan \alpha = \dfrac{1}{2}\). Tính \(\cot \alpha \)

$\sin {743^0} = \sin {23^0}$

$\sin {743^0} =  - \sin {23^0}$         

$\sin {743^0} = \cos {\rm{2}}{{\rm{3}}^0}$         

$\sin {743^0} =  - \cos {\rm{2}}{{\rm{3}}^0}$

\({33^0}45'.\)

\( - {29^0}30'.\)

\( - {33^0}45'.\)

\( - {32^0}55.\)

$\sin {225^0} = \dfrac{{ - \sqrt 2 }}{2}$

${\rm{cos22}}{{\rm{5}}^0} =  - \dfrac{{\sqrt 2 }}{2}$

$\tan {225^0} =  - 1$

$\cot {225^0} = 1$

\(\dfrac{{1 + m}}{2}\)

\(\dfrac{{1 - m}}{2}\)

\(\dfrac{{1 - m}}{5}\)

$m$

$\alpha $ và $\beta $; $\gamma $ và $\delta $.

\(\beta \) và $\gamma $; $\alpha $ và $\delta $.

$\alpha ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \beta ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \gamma $.

$\beta ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \gamma ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \delta $.

$\sin x > 0$    

$\cos x < 0$

$\tan x > 0$

$\cot x > 0$

\(\cot \alpha \tan \alpha  = 1,\quad \alpha  \ne \dfrac{{k\pi }}{2},k \in Z\)   

$1 + {\tan ^2}\alpha  = \dfrac{1}{{{{\cos }^2}\alpha }},{\rm{ }}\alpha  \ne {\rm{k}}\pi {\rm{, k}} \in Z$

${\sin ^2}\alpha  + c{\rm{o}}{{\rm{s}}^2}\beta  = 1$

$1 + {\cot ^2}\alpha  = \dfrac{1}{{{{\sin }^2}\alpha }},{\rm{ }}\alpha  \ne \dfrac{\pi }{2}{\rm{ + k}}\pi {\rm{, k}} \in Z$