Đáp án:
$\begin{array}{l}
\Delta ABC\,vuông\,tại\,A \Rightarrow A{B^2} + A{C^2} = B{C^2}\left( {pytago} \right)\\
\Rightarrow \sin B = \frac{{AC}}{{BC}};c{\rm{osB = }}\frac{{{\rm{AB}}}}{{BC}}\\
\Rightarrow {\sin ^2}B + co{s^2}B = {\left( {\frac{{AC}}{{BC}}} \right)^2} + {\left( {\frac{{AB}}{{BC}}} \right)^2} = \frac{{A{C^2} + A{B^2}}}{{B{C^2}}} = \frac{{B{C^2}}}{{B{C^2}}} = 1\\
\Rightarrow dpcm
\end{array}$
