√3tanAtanB-tanA-tanB
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解析:(1)√3*tanA*tanB-tanA-tanB=√3,也即是,tanA+tanB=-√3(1-tanA*tanB)故,tanC=-tan(A+B)=-(tanA+tanB)/(1-tanA*
1.倍角公式:cos2α=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)3+cos4a-4cos2a=3+(2cos^2(2a)-1)-4(1-2sin^2(a))
√3×tanAtanB-tanA-tanB=√3∴tanA+tanB=-√3(1-tanAtanB)∴tan(A+B)=(tanA+tanB)/(1-tanAtanB)=-√3∴tanC=tan[π-
(1)tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A+B)(1-tanAtanB)=tanA+tanB=根号3tanAtanB-根号3tan(A+B)=-根号3,tan(
tan(B+C)=(tanB+tanC)/(1-tanB*tanC)tanB+tanC+根号3tanBtanC=根号3,tanB+tanC=根号3-根号3tanBtanC=根号3*(1-tanB*ta
tanC=tan(180-(A+B))=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)tanA+tanB+3=3*tanAtanBtanA+tanB=3tanAtanB-3(t
由tanA+tanB+3=3tanAtanB可得tan(A+B)=tanA+tanB1-tanAtanB=-3因为A,B,C是三角形内角,所以A+B=120°,所以C=60°故答案为:60°
tanA+tanB=根号3*tanAtanB-根号3tanA+tanB=根号3(tanAtanB-1)tanA+tanB=-根号3(1-tanAtanB)(tanA+tanB)/(1-tanAtanB
(1)√3tanAtanB-tanA-tanB=√3√3tanAtanB-√3=tanA+tanB(tanA+tanB)/(1-tanAtanB)=-√3tan(A+B)=-√3∴A+B=120°∴C
tanA+tanB+√3=√3tanAtanBtanA+tanB=√3(tanAtanB-1)所以-√3=(tanA+tanB)/[1-tanAtanB]tanC=tan(180-A-B)=-tan(
sinBcosB=(√3)/42sinBcosB=(√3)/2sin2B=(√3)/22B=60°,B=30°tanA+tanB=√3tanAtanB-√3,tanA+tanB=-√3(1-tanAt
由公式:tanA=a,tanB=b,tan(A+B)=(a+b)/(1-a*b)所以tan(A+B)=-√3,所以A+B=120度,所以角C=60度S△ABC=0.5*ab*sinc所以ab=6cos
1)根号3*tanAtanB-tanA-tanB=根号3tanA+tanB=根号3(tanAtanB-1)tan(A+B)=(tanA+tanB)/(1-tanAtanB)=根号3*(tanAtanB
不相等,正确的式子应该是tan(A+B)=tanA+tanB+tanAtanBtan(A+B)推倒的方式如下:∵tan(A+B)=(tanA+tanB)/(1-tanAtanB)tanA+tanB=(
1(√3)(tanAtanB+a)+2tanA+3tanB=0①(tanA+tanB)/(1-tanAtanB)=tan(A+B)=1/(√3)3(tanA+tanB)+(√3)(tanAtanB-1
tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(A+B)(1-tanAtanB)=tanA+tanB=根号3tanAtanB-根号3tan(A+B)=-根号3,tan(180
tanA+tanB+√3=√3tanAtanBtanA+tanB=-√3(1-tanAtanB)tan(A+B)=(tanA+tanB)/(1-tanAtanB0=-√3A+B=120,C=60度再由
∵tanA+tanB+√3tanAtanB=根号3∴等式两边同÷根号3,得(tanA+tanB)/根号3+tanAtanB=1移项得(tanA+tanB)/根号3=1-tanAtanB,∴tanAta
/>1.已知tanA+tanB=√3tanAtanB-√3,即(tanA+tanB)/(1-tanAtanB)=-√3tan(A+B)=(tanA+tanB)/(1-tanAtanB);所以tan(A
tanB+tanC+√3tanBtanC=√3可以化为:tanB+tanC=√3(1-tanBtanC)(tanB+tanC)/(1-tanBtanC)=√3即tan(B+C)=√3因为B,C都为三角