Cho tam giác ABC. Chứng minh rằng
a) \(\cot A = \frac{{{b^2} + {c^2} - {a^2}}}{{4S}}\) (S là diện tích tam giác ABC);
b) \(\cot A + \cot B + \cot C = \frac{{{a^2} + {b^2} + {c^2}}}{{4S}}\)
Ta có
a)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}};\\
S = \frac{1}{2}bc.\sin A
\end{array}\\
\begin{array}{l}
\Rightarrow \cot A = \frac{{\cos A}}{{\sin A}} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc.\sin A}}\\
= \frac{{{b^2} + {c^2} - {a^2}}}{{4S}}
\end{array}
\end{array}\)
b) Tương tự câu a), ta có:
\(\begin{array}{*{20}{l}}
{\cot B = \frac{{{a^2} + {c^2} - {b^2}}}{{4S}};\cot C = \frac{{{a^2} + {b^2} - {c^2}}}{{4S}}}\\
\begin{array}{l}
\Rightarrow \cot A + \cot B + \cot C\\
= \frac{{{b^2} + {c^2} - {a^2}}}{{4S}} + \frac{{{a^2} + {c^2} - {b^2}}}{{4S}} + \frac{{{a^2} + {b^2} - {c^2}}}{{4S}}
\end{array}\\
{ = \frac{{{a^2} + {b^2} + {c^2}}}{{4S}}}
\end{array}\)
-- Mod Toán 10
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