Cho tam giác $ABC$ thỏa mãn :   $\frac{\sin ^2 A }{ \cos A} +\frac{\sin ^2B }{ \cos B} = (\sin A+\sin B)\cot \frac{ C}{ 2}    (1)$
Chứng minh $\Delta ABC$ cân.
$(1)   \Leftrightarrow \sin A \tan A + \sin B \tan B=\sin A \tan \frac{ A+B}{ 2} +\sin B \tan \frac{ A+B}{ 2} $
          $\Leftrightarrow \sin A \left ( \tan A - \tan \frac{A+B }{ 2} \right ) + \sin B \left ( \tan B - \tan \frac{ A+B}{ 2}  \right )=0$
          $\Leftrightarrow \frac{\sin A \sin \frac{ A-B}{ 2}  }{ \cos A \cos \frac{ A+B}{ 2} } +\frac{ \sin B \sin \frac{ B-A}{ 2} }{ \cos B \cos \frac{ A+B}{ 2} } =0$
          $\Leftrightarrow \sin \frac{ A-B}{ 2} (\tan A - \tan B)=0$
          $\Leftrightarrow \left[ \begin{array}{l}
\sin \frac{{A - B}}{2} = 0\\
\tan A = \tan B
\end{array} \right.  \Leftrightarrow  A = B$
Vậy $\Delta ABC$ cân đỉnh $C.$

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