A. \(\sin \frac{A}{2} = \cos \frac{B}{2}\cos \frac{C}{2} - \sin \frac{B}{2}\sin \frac{C}{2}\)
B. \(\frac{{{{\tan }^2}A - {{\tan }^2}B}}{{1 - {{\tan }^2}A{{\tan }^2}B}} = - \tan \left( {A - B} \right)\tan C\)
C. \(\cot A\cot B + \cot B\cot C + \cot C\cot A = 1\)
D. \({\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2} = 2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)
D
+ \(\sin \frac{A}{2} = \cos \left( {\frac{\pi }{2} - \frac{A}{2}} \right) = \cos \left( {\frac{{B + C}}{2}} \right) = \cos \frac{B}{2}\cos \frac{C}{2} - \sin \frac{B}{2}\sin \frac{c}{2}\)
+ \(\frac{{{{\tan }^2}A - {{\tan }^2}B}}{{1 - {{\tan }^2}A{{\tan }^2}B}} = \tan \left( {A + B} \right)\tan \left( {A - B} \right)\)
\( = - \tan \left( {A - B} \right)\tan \left( {\pi - A - B} \right) = - \tan \left( {A - B} \right)\tan C\)
+ \(\cot A\cot B + \cot B\cot C + \cot C\cot A\)
Có \(\cot C = \cot \left( {\pi - A - B} \right) = - \cot \left( {A + B} \right) = \frac{{1 - \cot A\cot B}}{{\cot A + \cot B}}\)
\( \Rightarrow \cot A\cot B + \cot C\left( {\cot A + \cot B} \right) = \cot A\cot B + 1 - \cot A\cot B\)
Vậy D sai.
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