已知a,b都是锐角,tana=1 7

cos(a+b)=-5/13,=>sin(a+b)=12/13,=>tan(a+b)=-12/5,tana=3/4,=>tan(a+b)=(tana+tanb)/(1-tana*tanb)=(3/4+

因为A,B都是锐角,所以sin(A+B)=5根号3/14,所以tan(A+B)=-5根号3/11,把tan(A+B)=-展开,可得tanB=根号3.又因为B为锐角,所以cosB=1/2!

因为A,B都是锐角,所以sin(A+B)=5根号3/14,所以tan(A+B)=-5根号3/11,把tan(A+B)=-展开,可得tanB=根号3.又因为B为锐角,所以cosB=1/2

tanA=tan[(B+A)-B]=[tan(B+A)-tanB]/[1+tan2BtanB]=tanB/[1+2(tanB)^2]=1/[1/tanB+2tanB]≤1/(2√2)=√2/4

∵：sinB=(√10)/10,∴：cosB=√[1-(sinB)^2]=(3√10)/10tanB=sinB/cosB=[(√10)/10]/[(3√10)/10]=1/3又：tanA=1/7tan

a,b为锐角,cosa=3/5,sina=4/5tana=4/3tan(a-b)=(tana-tanb)/(1+tanatanb)=(4/3-tanb)/(1+4tanb/3)(4/3-tanb)/(

cosAcosB-sinAsinB=sinAcosB-cosAsinBcosA(sinB+cosB)=sinA(sinB+cosB)因为B是锐角,所以sinB+cosB不等于0cosA=sinAtan

cos(a+b)=sin(a-b)cosacosb-sinasinb=sinacosb-cosasinbcosacosb+cosasinb=sinasinb+sinacosbcosa(cosb+sin

1.cos（A+B)=cosAcosB-sinAsinBsin（A-B)=sinAcosB-sinBcosAcosAcosB-sinAsinB=sinAcosB-sinBcosAcosA(cosB+s

cos2B=cosB^2-sinB^2=2/5;2B

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

tanA=√3*（1+m),tan（-B）=√3*(tanA+tanB+m)tanA-tan(-B)=√3(1-tanAtanB)=tanA+tanBtan(A+B)=[tanA+tanB]/[1-t

sina=4√3/7,cosa=1/7sin(a+b)=5√3/14cosb=cos[(a+b)-a]=.

等号两边拆开移项和并同类项约分得sinA=cosA所以tanA=1

1、tanAtanB=tanA＋tanB＋1tanAtanB－1=tanA＋tanB则：tan(A＋B)=[tanA＋tanB]/[1－tanAtanB]=－1因为A、B为锐角,则：A＋B=3π/4,

3tana=tan(a+b)=(tana+tanb)/(1-tanatanb)∴3tana-3tanatanb=tana+tanb∴(1+3tana)tanb=2tana∴tanb=2tana/(1+

tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(2+3)/(1-2*3)=5/(-5)=-1tan135=-tan45=-1即A+B=135度

SinA/SinB=Cos(A+B)SinA=Cos(A+B)SinB＝1/2[sin(2B+A)-sinA]3sinA=sin(2B+A)可见当sin(2B+A)＝1＝3sinA时sinA有最大值1

题目有问题cosA=3/545°而tan(A-B)=-1tan(B-A)=1B-A=45°B=A+45°>90°所以你算的tanB

是(sina+cosa)/(sina-cosa)还是sina+cosa/sina-cosa无括号?是(sina+cosa)/(sina-cosa)的话=[(sina+cosa)/cosa]/[(sin