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[Drago]

$\Delta  ABC $: $tan\frac{A}{2}+tan\frac{B}{2}= 1$

Chứng minh :$\frac{3}{4}\leq tan\frac{C}{2}< 1$

Edited by Drago, 02-05-2017 - 21:43.

[ThuThao36]

 

$\Delta  ABC $: $tan\frac{A}{2}+tan\frac{B}{2}= 1$

Chứng minh :$\frac{3}{4}\leq tan\frac{C}{2}< 1$

 

Đầu tiên chứng minh đẳng thức:

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

Từ đó suy ra:

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

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

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

$tan\frac{A}{2}.tan\frac{B}{2}\leq \frac{(tan\frac{A}{2}+tan\frac{B}{2})^{2}}{4}=\frac{1}{4}$

$\Rightarrow tan\frac{C}{2}\geq 1-\frac{1}{4}=\frac{3}{4}$

Từ (1) ta cũng suy ra $tan\frac{C}{2}< 1$