[Đại 10] lượng giác

D

dien0709

Chứng minh tam giác ABC đều khi:
$tanA+tanB+tanC=cot \dfrac{A}{2}+cot \dfrac{B}{2}+cot \dfrac{C}{2}$

$A+B=180^o-C$

$tanA+tanB=\dfrac{sin(A+B)}{cosAcosB}=\dfrac{2sinC}{cos(A-B)+cos(A+B)}$

$=>tanA+tanB=\dfrac{2sinC}{cos(A-B)-cosC}$

$\ge \dfrac{2sinC}{1-cosC}=\dfrac{4sin\dfrac{C}{2}cos\dfrac{C}{2}}{2sin^2\dfrac{C}{2}}=2cot\dfrac{C}{2}$

Tương tự $tanB+tanC\ge 2cot\dfrac{A}{2} , tanC+tanA\ge 2cot\dfrac{B}{2}$

Cộng lại=>$tanA+tanB+tanC\ge cot\dfrac{A}{2}+cot\dfrac{B}{2}+cot\dfrac{C}{2}$

dấu"=" <=> A=B=C=>đpcm
 
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