Cho góc $x$  thoả ${0^0} < x < {90^0}$ . Trong các mệnh đề sau, mệnh đề nào sai:

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

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

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

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

$A =  - 3 - \sqrt 3$

$A = 2 - 3\sqrt 3$

$A = \dfrac{{2\sqrt 3 }}{{\sqrt 3  - 1}}$

$A = \dfrac{{1 - \sqrt 3 }}{{\sqrt 3 }}$

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

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

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

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

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

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

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

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

$n-p$

$m + p$

$m-p$

$n + p$

$\dfrac{1}{2}$

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

$1$

$3$