Cho tam giác \(ABC\), chứng minh rằng:
a) \(\tan A + \tan B + \tan C = \tan A{\rm{ }}{\rm{. }}\tan B{\rm{ }}{\rm{. }}\tan C\)
(với điều kiện tam giác \(ABC\) không vuông)
b) \(\tan \frac{A}{2}{\rm{ }}{\rm{. }}\tan \frac{B}{2} + \tan \frac{B}{2}{\rm{ }}{\rm{. }}\tan \frac{C}{2} + \tan \frac{C}{2}{\rm{ }}{\rm{. }}\tan \frac{A}{2} = 1\)
Sử dụng định lý tổng ba góc trong một tam giác: \(A + B + C = \pi \)
Sử dụng công thức \(\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\)
Trong tam giác \(ABC\), ta có \(A + B + C = \pi \).
a) Do \(A + B + C = \pi \Rightarrow A + B = \pi - C \Rightarrow \tan \left( {A + B} \right) = \tan \left( {\pi - C} \right)\)
Advertisements (Quảng cáo)
Vì \(\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\), \(\tan \left( {\pi - C} \right) = \tan \left( { - C} \right) = - \tan C\), nên:
\(\tan \left( {A + B} \right) = \tan \left( {\pi - C} \right) \Rightarrow \frac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C\)
\( \Rightarrow \tan A + \tan B = - \left( {1 - \tan A\tan B} \right)\tan C\)
\( \Rightarrow \tan A + \tan B = - \tan C + \tan A\tan B\tan C \Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C\)
Bài toán được chứng minh.
b) Ta có:
\(A + B + C = \pi \Rightarrow \frac{{A + B + C}}{2} = \frac{\pi }{2} \Rightarrow \frac{{A + B}}{2} = \frac{\pi }{2} - \frac{C}{2} \Rightarrow \tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\)Do \(\tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}}\) và \(\tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right) = \cot \frac{C}{2} = \frac{1}{{\tan \frac{C}{2}}}\), nên:
\(\tan \left( {\frac{A}{2} + \frac{B}{2}} \right) = \tan \left( {\frac{\pi }{2} - \frac{C}{2}} \right) \Rightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \frac{1}{{\tan \frac{C}{2}}}\)
\( \Rightarrow \left( {\tan \frac{A}{2} + \tan \frac{B}{2}} \right)\tan \frac{C}{2} = 1 - \tan \frac{A}{2}\tan \frac{B}{2} \Rightarrow \tan \frac{A}{2}\tan \frac{B}{2} + \tan \frac{B}{2}\tan \frac{C}{2} + \tan \frac{C}{2}\tan \frac{A}{2} = 1\)
Bài toán được chứng minh.