Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{S_{ABC}} = \dfrac{1}{2}bc.\sin A = \dfrac{1}{2}ca.\sin B = \dfrac{1}{2}ab.\sin C\\
\Rightarrow \sin A = \dfrac{{2S}}{{bc}};\,\,\,\,\sin B = \dfrac{{2S}}{{ca}};\,\,\,\,\sin C = \dfrac{{2S}}{{ab}}\\
\cot A + \cot B + \cot C\\
= \dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos B}}{{\sin B}} + \dfrac{{\cos C}}{{\sin C}}\\
= \dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{{\dfrac{{2S}}{{bc}}}} + \dfrac{{\dfrac{{{c^2} + {a^2} - {b^2}}}{{2ca}}}}{{\dfrac{{2S}}{{ca}}}} + \dfrac{{\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{{\dfrac{{2S}}{{ab}}}}\\
= \dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{2}}}{{2S}} + \dfrac{{\dfrac{{{c^2} + {a^2} - {b^2}}}{2}}}{{2S}} + \dfrac{{\dfrac{{{a^2} + {b^2} - {c^2}}}{2}}}{{2S}}\\
= \dfrac{{{b^2} + {c^2} - {a^2}}}{{4S}} + \dfrac{{{c^2} + {a^2} - {b^2}}}{{4S}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{4S}}\\
= \dfrac{{{a^2} + {b^2} + {c^2}}}{{4S}}
\end{array}\)
