若tana-tanc=根号3 3(1 tanatanc)

(a+b+c)(a-b+c)=3ac,(a+c)^2-b^2=3ac,a^2+c^2-b^2=ac,cosB=(a^2+c^2-b^2)/2ac=1/2,B=π/3=60度.tan(A+C)=-tan

∵A+B＝π-C,∴tan(A+B)=tan(π-C)即：(tanA+tanB)/(1-tanA*tanB)=-tanC,∴tanA+tanB=-tanC(1-tanAtanB)即：tanA+tanB

ABC分别是三角形内角,2B=A+CtanB=tan（A/2+C/2）=（tanA/2+tanC/2）/（1-tanA/2*tanC/2）=√3所以tanA/2+tanC/2+√3tanA/2tanC

∵tan(A+B)=tanA+tanB/1-tanA*tanBtan(A+B)=tan(π-C)=-tanC∴tanA+tanB/1-tanA*tanB=-tanC整理移项即得tanA+tanB+ta

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

tanA+tanB+tanC=tan(A+B)(1-tanAtanB)+tanC=tan(pai-c)(1-tanAtanB)+tanC=-tanC(1-tanAtanB)+tanC=tanAtanB

tanA+tanc=tan(A+C)(1-tanAtanC)又因为tan(A+C)=-tan(B）根据tan(180-a)=-tana所以tanA+tanB+tanC=tanB-tanB(1-tanA

（1）2B=A+C得到B=60tan120=tan(A+C)=（tanA+tanC）/（1-tanA*tanC）=-根号3乘过来移项得到tanA+tanC-根号3tanA乘tanC的值是根号3（2）t

输入有误吧tan²B=tanAtanC,tan(A+C)=(tanA+tanC)/(1-tanAtanC)=(3√3-tanB)/(1-tan²B)所以-tanB=(3√3-tan

B=60度A+C=120度tan(A+C)=-√3=(tanA+tanC)/(1-tanAtanC)=(3+√3)/(1-tanAtanC)--->1-tanAtanC=(3+√3)/(-√3)=-√

应该是在三角形中吧三角形中A+B+C=3.143.14-A=B+CtanA=-tan(3.14-A)=-tan(B+C)=(tanB+tanC)/(tanBtanC-1)所以tanA(tanBtanC

∵tan(A+B)=[tanA+tanB]/[1-tanA*tanB]tan(A+B)=tan(π-C)=-tanC∴tanA+tanB/1-tanA*tanB=-tanC整理移项即得tanA+tan

tanb=tan(180-a-c)=-tan(a+c)=-(tana+tanc)/(1-tanga*tanc)因为tana+tanb+tanc=3倍根号下3,所以tanb=-(3倍根号下3-tanb)

cosB=3*√10/10sinB=√10/10tanB=1/3tanA=1/2tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=(5/6)/(1-1/6)=-1

∵tan(A+B)=tanA+tanB/1-tanA*tanBtan(A+B)=tan(π-C)=-tanC∴tanA+tanB/1-tanA*tanB=-tanC整理移项即得tanA+tanB+ta

(1+tanC/tanA)+(1+tanC/tanB)=2+tanC/tanA+tanC/tanB=6则tanC/tanA+tanC/tanB=4

tanB+tanC=-√3(1-tanBtanC)tan(B+C)=(tanB+tanC)/(1-tanBtanC)=-√3tan(180-A)=-tanA=-√3tanA=√3A=60度√3(tan

应该求tanB+tanC吧!由sinA=sin[∏-(B+C)]=即sin(B+C)=2cosBcosC,展开得,sinBcosC+sinCcosB=2cosBcosC,sinBcosC-cosBco

tanB+tanC+(根号3)*tanBtanC=根号3移项可得:左边=tanB+tanC=根号3-(根号3)*tanBtanC=根号3(1-tanBtanC)=根号3*(tanB+tanC)/tan

a+b+c)(a-b+c)=3ac,(a+c)^2-b^2=3ac,a^2+c^2-b^2=ac,cosB=(a^2+c^2-b^2)/2ac=1/2,B=π/3=60度.tan(A+C)=-tanB