Cho tam giác ABCthỏa sin2A+sin2B+sin2C=2. Chứng minh tam giác ABC là tam giác vuông.

\[\begin{array}{l}\,\,\,\,\,\,{\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 2\\ \Leftrightarrow \frac{{1 - \cos 2A}}{2} + \frac{{1 - \cos 2B}}{2} + {\sin ^2}C = 2\\ \Leftrightarrow 2 - \cos 2A - \cos 2B + 2{\sin ^2}C = 4\\ \Leftrightarrow \cos 2A + \cos 2B + 2{\cos ^2}C = 0\\ \Leftrightarrow 2\cos (A + B).\cos (A - B) + 2{\cos ^2}C = 0\\ \Leftrightarrow  - 2\cos C.\cos (A - B) + 2{\cos ^2}C = 0\\ \Leftrightarrow \cos C.[ - \cos (A - B) + \cos C] = 0\\ \Leftrightarrow \cos C.[\cos (A - B) + \cos (A + B)] = 0\\ \Leftrightarrow \cos C.\cos A.\cos B = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos A = 0\\\cos B = 0\\\cos C = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}A = \frac{\pi }{2}\\B = \frac{\pi }{2}\\C = \frac{\pi }{2}\end{array} \right.\end{array}\]