Cho tam giác ABC, chứng minh rằng: a) tan A + tan B + tan C = tan A

Bài 29 trang 16 SBT Toán 11 Tập 1: Cho tam giác ABC, chứng minh rằng:

a) tan A + tan B + tan C = tan A . tan B . tan C (với điều kiện tam giác ABC không vuông);

b) $\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}=1$ .

Trả lời

a) Vì tam giác ABC không vuông nên A, B, C khác $\frac{\pi }{2}$ , do đó tan A, tan B, tan C xác định.

Do A + B + C = π nên A + B = π – C, do đó tan(A + B) = tan(π – C) = tan(– C) = – tanC.

Mà $\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$ .

Khi đó $\frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$

⇔ tan A + tan B = – tan C . (1 – tan A . tan B)

⇔ tan A + tan B = – tan C + tan A . tan B . tan C

⇔ tan A + tan B + tan C = tan A . tan B . tan C.

b) Ta có $\frac{A+B+C}{2}=\frac{\pi }{2}$ , suy ra $\frac{A}{2}+\frac{B}{2}=\frac{\pi }{2}-\frac{C}{2}$  nên $\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \frac{C}{2}$

$\iff \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}$

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

$\iff \tan \frac{A}{2}.\tan \frac{C}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{A}{2}.\tan \frac{B}{2}=1$

$\iff \tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}=1$.