Cho A, B, C là ba góc của một tam giác không vuông. Hệ thức nào sau đây SAI ?

* Hướng dẫn giải

Ta có

\(\cos \frac{B}{2}\cos \frac{C}{2} - \sin \frac{B}{2}\sin \frac{C}{2} = \cos \left( {\frac{B}{2} + \frac{C}{2}} \right) = \cos \left( {\frac{\pi }{2} - \frac{A}{2}} \right) = sin\frac{A}{2}.\) A đúng

\(\tan A + \tan B + \tan C = \tan A.\tan B.\tan C \Leftrightarrow  - \tan A\left( {1 - \tan B\tan C} \right) = \tan B + \tan C\)

\( \Leftrightarrow \tan A =  - \frac{{\tan B + \tan C}}{{1 - \tan B\tan C}} \Leftrightarrow \tan A =  - \tan \left( {B + C} \right)\) B đúng

\(\cot A + \cot B + \cot C = \cot A.\cot B.\cot C \Leftrightarrow \cot A\left( {\cot B\cot C - 1} \right) = \cot B + \cot C\)

\( \Leftrightarrow \frac{1}{{\cot A}} = \frac{{\cot B\cot C - 1}}{{\cot B + \cot C}} \Leftrightarrow \tan A = \cot \left( {B + C} \right).\)

\(\tan \frac{A}{2}.\tan \frac{B}{2} + \tan \frac{B}{2}.\tan \frac{C}{2} + \tan \frac{C}{2}.\tan \frac{A}{2} = 1 \Leftrightarrow \tan \frac{A}{2}.\left( {\tan \frac{B}{2} + \tan \frac{C}{2}} \right) = 1 - \tan \frac{B}{2}.\tan \frac{C}{2}\)

\( \Leftrightarrow \frac{1}{{\tan \frac{A}{2}}} = \frac{{\tan \frac{B}{2} + \tan \frac{C}{2}}}{{1 - \tan \frac{B}{2}.\tan \frac{C}{2}}} \Leftrightarrow \cot \frac{A}{2} = \tan \left( {\frac{B}{2} + \frac{C}{2}} \right)\) D đúng