tanC=(simA sinB) (cosA cosB),sin(B-A)

∵A+B＝π-C,∴tan(A+B)=tan(π-C)(tanA+tanB)/(1-tanA*tanB)=-tanC,tanA+tanB=-tanC+tanAtanBtanC∴tanA+tanB+ta

∵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

tanA=-tan(B+C)=-(tanB+tanC)/(1-tanBtanC)由均值不等式,3=tanB+tanC>=2根号下（tanBtanC）所以tanBtanC=-3/(1-9/4)=12/5

∵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=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/

tan[180-(-c)]=-tan（-c）＝tanc也就是tan(180＋c)=tanc这个肯定是正确的.

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

tanB/2=tan(π-A-C)/2=tan[π/2-(A+C)/2]=cot(A+C)/2=(1-tanA/2*tanC/2)/(tanA/2+tanC/2)因此tanA/2tanB/2+tanB

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

把tan换做sin/cos,再用正弦定理,余弦定理代入就能反证了.sinA/cosA+sinB/cosB+sinC/cosC=sinAsinBsinC/cosAcosBcosC正弦定理得：a/cosA

∵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

三角形中,A+B+C=180°∴A+B=180°-C∴tan(A+B)=tan(180-C)∴tan(A+B)=-tanC

∵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

tan(b+c)=(tanb+tanc)/(1-tanbtanc)tan(b+c)(1-tanbtanc)=tanb+tanctan(b+c)-tanbtanctan(b+c)=tanb+tanc所以

A/2+B/2+C/2=90°A/2=90°-(B/2+C/2)tanA/2=tan(90°-(B/2+C/2))=cot(B/2+C/2)=1/tan(B/2+C/2)=(1-tanB/2tanC/

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

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