已知tana =7分之1,tanb=3分之1

tanA加tan分之1等于3分之4倍根号3tan²A+1=4√3/3tanAtan²A-4√3/3tanA+1=0(tanA-√3)(tanA-√3/3)=0tanA=√3或tan

tan(a-b)=(tana-tanb)/(1+tana*tanb)因为tana=3所以tan(a-4分之派）=(tana-4分之派)/(1+tana*4分之派)=(3-1)/(1+3*1)=1/2

tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+si

因为tanb＝1/3所以tan（2b）＝2tanb/（1－tanbtanb）＝3/4,所以tan（a＋2b）＝（tana＋tan2b）/（1－tanatan2b）,将tana＝1/7和tan2b＝3/

tanA=sinA/cosA所以tanA+1/tanA=sinA/cosA+cosA/sinA=(sinA^2+cosA^2)/sinAcosA又因为sina^2+cosA^2=1所以原式等于1/si

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

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

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

tan(π/4＋a)=[tan(π/4)＋tana]/[1－tan(π/4)tana]=(1＋tana)/(1－tana)=1/2解得：tana=－1/3

tan(a+β)=7(tana+tanβ)/(1-tanatanβ)=73(tana+tanβ)=73sin(a+β)/cosacosβ=7cosacosβ=3sin(a+β)/7=3(7/5√2)/

tan(a+π/4)=(tana+tan(π/4))/[1-tana*tan(π/4)]=(3+1)/(1-3*1)=-2没有疑问请采纳为满意答案,再答：没有疑问请采纳为满意答案，谢谢。

结果:tana*tanb=1/2.过程也不复杂,把tana移项,然后展开tan(a+b),再全部通分,两边合并同类项.

证明：tan(a+b)=(tana+tanb)/(1-tana·tanb)∴tan(a+π/4)=[tana+tan(π/4)]/[1-tana·tan(π/4)]=(1+tana)/(1-tana)

分析法倒推tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(

tan3a=[3tana-(tana)^3]/[1-3(tana)^2]=1/tana,∴3(tana)^2-(tana)^4=1-3(tana)^2,∴(tana)^4-6(tana)^2+1=0,

tan(a+2b)=(tana+tan2b)/(1-tanatan2b)tan2b=2tanb/(1-tan^2b)tana=1/7,tanb=1/3带入tan2b=2tanb/(1-tan^2b)=

tan(a+b)=(-7-3)/(21-1)=-1/2tan(2a+b)=[tana+tan(a+b)]/[1-tana*tan(a+b)]=(-2-3)/(6-1)=-1

tan(a+2b)=(tana+tan2b)/(1-tanatan2b)tan2b=2tanb/(1-tan^2b)tana=1/7,tanb=1/3带入tan2b=2tanb/(1-tan^2b)=

∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6

应该是tan(π/4-a)=(tanπ/4-tana)/(tanπ/4+tana)=(1-tana)/(1+tana)=根号5