Cho tam giác ABC. Chứng minh: tan ( A/2)tan ( B/2) + tan ( B/2)tan ( C/2) + tan ( C/2)tan( A/2) = 1
Câu hỏi:
Cho tam giác ABC. Chứng minh:
\(\tan \left( {\frac{A}{2}} \right)\tan \left( {\frac{B}{2}} \right) + \tan \left( {\frac{B}{2}} \right)\tan \left( {\frac{C}{2}} \right) + \tan \left( {\frac{C}{2}} \right)\tan \left( {\frac{A}{2}} \right) = 1\).
Trả lời:
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).