已知a²tanB=b²tanA,试判断

右边=tan(a+b)[1-tana*tanb]=[(tana+tanb)/(1-tana*tanb)]/[1-tana*tanb]=tana+tanb=左边

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

1.tanA-tanB/tanA+tanB=c-b/c是不是(tanA-tanB)/(tanA+tanB)=(c-b)/c?是的话,现在就解吧.假如是(tanA-tanB)/(tanA+tanB)=(

将tan(a+b)化简,易知tana*tanb=1/2

LZ,∠A=60度.\x0d\x0d(tanA-tanB)/(tanA+tanB)=1-2tanB/(tanA+tanB)\x0d(c-b)/c=1-b/c\x0d由已知可得,\x0d2tanB/(t

∠A=60度.(tanA-tanB)/(tanA+tanB)=1-2tanB/(tanA+tanB)(c-b)/c=1-b/c由已知可得,2tanB/(tanA+tanB)=b/c=sinB/sinC

tanA+tanB=5,tanA*tanB=6可解得tanA=3,tanB=2（因为a>b）从而有sinA=3/√10,sinB=2/√5tanC=-tan(A+B)=-(tanA+tanB)/(1-

tan(a+b)=4(tana+tanb)/(1-tanatanb)=4tana+tanb=2(1)所以2/(1-tanatanb)=4所以tanatanb=1/2(2)由（1）（2）tana,tan

tan(A+B)=4(tanA+tanB)/(1-tanAtanB)=4tanAtanB=1/2再联立tanA+tanB=2又tanA

1、tan(A－B)=[tanA－tanB]/[1＋tanAtanB]=√3/3,A－B=30°,C=90°；2、|m|=|n|=1,|3m－2n|²=13－12sin(A＋B)=13－12

tanA=根号3(1+m)根号3(tanA*tanB+m)+tanB=0所以：3(1+m)tanB+根号3m+tanB=0(4+3m)tanB=-根号3m所以：tanB=-根号3m/(4+3m)因为t

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

tanA/tanB=sinAcosB/sinBcosAc=2RsinCb=2RsinB所以2x2RsinC-2RsinB/2RsinB=2sinC-sinB/sinB所以sinAcosB/sinBco

（tana-tanb）/（tana+tanb）=（sina/cosa-sinb/cosb）/（sina/cosa+sinb/cosb）=(sinacosb-cosasinb)/(sinacosb+co

(tanA+tanB)/(1-tanA*tanB)=-1两边同乘以(1-tanA*tanB),等式两边就为(tanA+tanB)=-(1-tanA*tanB),“-“(1-tanA*tanB)注意这个

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=(

tan（a+b）=(tana+tanb)/(1-tanatanb)4=2/(1-tanatanb)所以tanatanb=1/2

tan(A+B)=(tanA+tanB)/(1-tanA*tanB)

tan(A+B)=4=(tanA+tanB)/(1-tanAtanB)tanAtanB=1/2tanA

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