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Ta có:
$\cot 2C= \frac{1}{2}(\cot C - \cot B)$
$\Leftrightarrow \frac{1-\tan^2C}{2\tan C}=\frac{1}{2}(\cot C - \cot B)$
$\Leftrightarrow \frac{1}{2}(\cot C - \tan C)=\frac{1}{2}(\cot C - \cot B)$
$\Leftrightarrow \tan C=\cot B$
$\Leftrightarrow B+C=\frac{\pi}{2}\Leftrightarrow \Delta ABC$ vuông tại $A$.
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