Cho $A = \cos {235^0}.\sin {60^0}.\tan {125^0}.\cos {90^0}{\rm{ }}$. Khẳng định nào sau đây đúng?

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

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

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

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

\(A =  - 3 - \sqrt 3 \)

\(A = 2 - 3\sqrt 3 \)

\(A = \dfrac{{2\sqrt 3 }}{{\sqrt 3  - 1}}\)

\(A = \dfrac{{1 - \sqrt 3 }}{{\sqrt 3 }}\)

$\sin x > 0$    

$\cos x < 0$

$\tan x > 0$

$\cot x > 0$

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

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

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

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

$\dfrac{{ - 5}}{{13}};{\rm{ }}\dfrac{2}{3}$

$\dfrac{2}{3};\dfrac{{ - 5}}{{12}}$

$\dfrac{{ - 5}}{{13}};{\rm{ }}\dfrac{5}{{12}}$  

$\dfrac{5}{{13}};{\rm{ }}\dfrac{{ - 5}}{{12}}$

$n-p$

$m + p$

$m-p$

$n + p$

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

$\dfrac{{ - 1}}{2}$

\(1\)

\(3\)