Trong mặt phẳng tọa độ \(Oxy,\) cho tam giác \(ABC\) có \(A\left( {10;5} \right),{\rm{ }}B\left( {3;2} \right)\) và \(C\left( {6; - 5} \right)\). Khẳng định nào sau đây là đúng?

\({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\).

\(1 + {\cot ^2}\alpha = \dfrac{1}{{{{\sin }^2}\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {\sin \alpha \ne 0} \right)\).

\(\tan \alpha .\cot \alpha  =  - 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {\sin \alpha .\cos \alpha  \ne 0} \right)\)

\(1 + {\tan ^2}\alpha = \dfrac{1}{{{{\cos }^2}\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {\cos \alpha \ne 0} \right)\).

\({\sin ^2}\alpha  + \cos {\alpha ^2} = 1\).

\({\sin ^2}\alpha  + {\cos ^2}\dfrac{\alpha }{2} = 1\).

\(\sin {\alpha ^2} + \cos {\alpha ^2} = 1\).

\({\sin ^2}2\alpha  + {\cos ^2}2\alpha  = 1\).

$25$

$5$     

$6$     

$7$

\(bc = 2R.{h_a}\)

\(ac = R.{h_b}\)

\({a^2} = R.{h_a}\)

\(ab = 4R.{h_c}\)

\({a^2} = {b^2} + {c^2} - bc\cos A\)           

\({a^2} = {b^2} + {c^2} + bc\cos A\)

${a^2} = {b^2} + {c^2} - 2bc\cos A$          

\({a^2} = {b^2} + {c^2} + 2bc\cos A\)