在三角形abc种tanc=3根号7

把tan换做sin和cos,在用余旋定理

∵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

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

(1)tanC=3√7>0,C为锐角,sinC=3√7cosC,(sinC)^2=63(cosC)^2.(sinC)^2+(cosC)^2=64(cosC)^2=1,cosC=1/8.(2)ab=5/

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

因为三角形ABC为锐角所以tanC＝tan[∏－（A＋B）]即tanC＝－（tanA＋tanB）÷（1－tanA×tanB）－tanC＝（tanA＋tanB）÷（1－tanA×tanB）－tanC＋t

tanC=tan(180-(A+B))=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)tanA+tanB+3=3*tanAtanBtanA+tanB=3tanAtanB-3(t

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

S三角形ABC=1/2*AB*BC*sinB=√3/4AB=√3,即AB=4作AD⊥BC交BC延长线于D,AD=2√3,BD=1/2AB=2,即CD=1,所以tanC=AD/CD=-2√3

解cosA=3/5∵A∈(0,π)∴sinA=4/5∴tanA=4/3tan[A+(B-A)]=[tanA+tan(B-A)]/[1-tanAtan(B-A)]=(4/3+1/2)/(1-2/3)=(

（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

在三角形ABC中已知三个边abc成等比数列因为tanA•tanC=(tanB)^2,设公比为q,tanA=tanB/q,tanC=q*tanB由tanB=-tan(A+C)=(tanA+t

cosA=4/5;sina=3/5;c=3*2/2/sina=5;a/sinA=b/sinB=c/sin/C=b/sin(A+C)sin(A+C)/sinC=b/c=2/5;2/5=sin(A+C)/

∵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

tanC=sinC/cosC=(sinA+sinB)/(cosA+cosB),交叉相乘得sin(C-A)=sin(B-C),故C-A=B-C或C-A+B-C=π,显然C-A=B-C成立,2C=A+B,

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)

∵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

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

tanC=tan[180-(A+B)]=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-7

tanC/tanA+tanC/tanB=1tanBtanC+tanAtanC=tanAtanBtanC(tanA+tanB)=tanAtanBsinC/cosC(sinA/cosA+sinB/cosB