Cho A: B; C là các góc nhọn và tanA = 1/2, tanB = 1/5, tanC = 1/8

* Hướng dẫn giải

Chọn C.

Áp dụng công thức cộng ta có:

$$\begin{gathered} \tan \left(A+B+C\right)=\frac{\tan \left(A+B\right)+\tan C}{1-\tan \left(A+B\right).\tan C}=\frac{{\displaystyle \frac{\tan A+\tan B}{1-\tan A.\tan B}}+\tan C}{1-{\displaystyle \frac{\tan A+\tan B}{1-\tan A.\tan B}}.\tan C} \\ =\frac{\tan A+\tan B+\tan C-\tan A.\tan B.\tan C}{1-\tan A.\tan B-\tan A.\tan C-\tan B.\tan C}=\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{5}}.{\displaystyle \frac{1}{8}}}{1-{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{2}}.{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{5}}.{\displaystyle \frac{1}{8}}}=\frac{{\displaystyle \frac{13}{16}}}{{\displaystyle \frac{13}{16}}}=1 \\ = \end{gathered}$$

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