Cho tam giác ABC. Chứng minh:

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

Lời giải

Ta có: \[\frac{A}{2} + \frac{B}{2} = \frac{\pi }{2} - \frac{C}{2}\]

\( \Rightarrow \tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\)

\( \Leftrightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \cot g\frac{C}{2}\)

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

\[ \Leftrightarrow \tan \frac{A}{2}\tan \frac{C}{2} + \tan \frac{B}{2}\tan \frac{C}{2} = 1 - \tan \frac{A}{2}\tan \frac{B}{2}\]

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