Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cot A + \cot C = 2\cot B\\
\frac{{\cos A}}{{\sin A}} + \frac{{\cos C}}{{\sin C}} = 2\frac{{\cos B}}{{\sin B}}\\
\Leftrightarrow \frac{{\cos A.\sin C + \sin A.\cos C}}{{\sin A.\sin C}} = \frac{{2\cos B}}{{\sin B}}\\
\Leftrightarrow \frac{{\sin \left( {A + C} \right)}}{{ - \frac{1}{2}\left[ {\cos \left( {A + C} \right) - \cos \left( {A - C} \right)} \right]}} = \frac{{2\cos B}}{{\sin B}}\\
\Leftrightarrow \frac{{\sin \left( {180^\circ - \left( {\widehat A + \widehat C} \right)} \right)}}{{ - \frac{1}{2}\left[ { - \cos \left( {180 - \left( {A + C} \right)} \right) - \cos \left( {A - C} \right)} \right]}} = \frac{{2\cos B}}{{\sin B}}\\
\Leftrightarrow \frac{{\sin B}}{{\frac{1}{2}\left( {\cos B + \cos \left( {A - C} \right)} \right)}} = 2\frac{{\cos B}}{{\sin B}}\\
\Leftrightarrow {\sin ^2}B = {\cos ^2}B + \cos B.\cos \left( {A - C} \right)\\
\Leftrightarrow \cos B.\cos \left( {A - C} \right) = {\sin ^2}B - {\cos ^2}B\\
\Leftrightarrow \cos \left( {A - C} \right) = 1\\
\Leftrightarrow A - C = 0\\
\Leftrightarrow \widehat A = \widehat C = \frac{{180^\circ - \widehat B}}{2} = 60^\circ = \widehat B
\end{array}\)
Vậy tam giác ABC là tam giác đều.
