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 $\delta$.

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

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

$\sin x > 0$    

$\cos x < 0$

$\tan x > 0$

$\cot x > 0$

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

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

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

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