Bài tập 29 trang 16 SBT Toán 11 Tập 1 Cánh diều - CD

Trong tam giác \(ABC\), ta có \(A + B + C = \pi \).

a) Do \(A + B + C = \pi \Rightarrow A + B = \pi - C\)

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

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

\(\Rightarrow \frac{{\tan \frac{A}{2} + \tan \frac{B}{2}}}{{1 - \tan \frac{A}{2}\tan \frac{B}{2}}} = \frac{1}{{\tan \frac{C}{2}}}\)

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

-- Mod Toán 11 HỌC247