Chứng minh tam giác ABC cân biết tam giác ABC thỏa \(\frac{{{{\sin }^2}A}}{{\cos A}} + \frac{{{{\sin }^2}C}}{{\cos C}} = (\sin A + \sin C)\cot \frac{B}{2}\).

* Hướng dẫn giải

Ta có : \(\frac{{{{\sin }^2}A}}{{\cos A}} + \frac{{{{\sin }^2}B}}{{\cos B}} = (\sin A + \sin B)\cot \frac{C}{2}\)

\(\begin{array}{l}
 \Leftrightarrow \sin A\tan A + \sin B\tan B = \sin A.\tan \left( {\frac{{A + B}}{2}} \right) + \sin B.\tan \left( {\frac{{A + B}}{2}} \right)\\
 \Leftrightarrow \sin A\left( {\tan A - \tan \left( {\frac{{A + B}}{2}} \right)} \right) + \sin B\left( {\tan B - \tan \left( {\frac{{A + B}}{2}} \right)} \right) = 0\\
 \Leftrightarrow \sin A\frac{{\sin \left( {\frac{{A - B}}{2}} \right)}}{{\cos A\cos \left( {\frac{{A + B}}{2}} \right)}} + \sin B\frac{{\sin \left( {\frac{{B - A}}{2}} \right)}}{{\cos B\cos \left( {\frac{{A + B}}{2}} \right)}} = 0\\
 \Leftrightarrow \tan A\sin \left( {\frac{{A - B}}{2}} \right) - \tan B\sin \left( {\frac{{A - B}}{2}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan A = \tan B\\
\sin \left( {\frac{{A - B}}{2}} \right)
\end{array} \right. \Leftrightarrow A = B
\end{array}\) 

Vậy tam giác ABC cân tại C.