已知 2sin^2a-1 =tan45

要证3sinB=sin(2A+B)即证3sin(A+B-A)=sin(A+B+A)即证3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA即证2s

tan^2θ+1=2（tan^2a+1）sec^2θ=2sec^2acos^2a=2cos^2θ所以cos2θ=-2sin^2acos2θ+2sin^2a=0

(tan(a+β)-tana-tanβ)/tan²βtan(a+β)=[tan(a+β)-t(a+β)(1-tana*tanβ)]/tan²βtan(a+β)=tan(a+β)[1

tan[b－(π/4)]=[tanb－tan(π/4)]/[1＋tan(π/4)tanb]=[tanb－1]/[1＋tanb]=1/4则：tanb=5/3,又：tan(a＋b)=2/5则：tana=t

tanA=sinA/cosA=2则：sinA=2cosA且：(cosA)^2=1/[1+(tanA)^2]=1/5故：(sinA)^2-3sinAcosA+1=(2cosA)^2-6(cosA)^2+

由tana=2且sina^2+cosa^2=1所以sina=2/√5,cosa=1/√5所以sina*cosa=2/5

sina/cosa=2sin^2a/cos^2a=4cos^2=1-sin^2a带入得sin^2a=0.8然后cosa/sina=0.5左右同乘以sin^2a得到sinacosa=0.4所以是1.2

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

左边将tan(a)和cot(a)分别化成sin(a)和cos(a)直接利用公式1+tan^2(a)=1/cos^2(a)和sin^2(a)+cos^2(a)=1就可以证明,详细过程你最好自己来证.

tan^2A=2tan^B+1(sinA/cosA)^2=2(sinB/cosB)^2+1sin^2A/cos^2A=2sin^2B/cos^2B+1sin^2A*cos^2B=2sin^2Bcos^

(4sina-2cosa)/(5cosa+3sina)=(4tana-2)/(5+3tana)=(-4/3-2)/[5+3(-1/3)]=(-10/3)/4=-10/12=-5/6

由tan²a=2tan²b+1得sin²a/cos²a=2sin²b/cos²b+1sin²acos²b=2sin

原式=（sina/cosa＋1）/（1-sina/cosa）=（tana＋1）/（1-tana）=-3

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(

证明：sin^2B=2sin^2A-1得cos^2B=1-sin^2B=1-(2sin^2A-1)=2(1-sin^2A)=2cos^2A于是tan^2A-2tan^B=sin^2A/cos^2A-2

sin(a+b)=sinacosb+cosasinb=1/2sin(a-b)=sinacosb-cosasinb=1/3所以sinacosb=(1/2+1/3)/2=5/12cosasinb=(1/2

cos(a/2)-sin(a/2)=1/5【cos(a/2)-sin(a/2)】²=(1/5)²cos²(a/2)-2cos(a/2)sin(a/2)+sin²

=sin²A-1-3sinAcosA=-cos²A-3sinAcosA=-(cos²A+3sinAcosA)/(sin²A+cos²A)=-(1+3t

cos[(a＋b)＋a]=3sin[(a＋b)－a]cos(a＋b)cosa－sin(a＋b)sina=3sin(a＋b)cosa－3cos(a＋b)sinacos(a＋b)[cosa＋3sina]=

由Sin（a-（2n+1）π/2）=3/5化简得cosa=3/5或-3/5故易知tana=4/3或-4/3所以tana+1/tana=25/12或-25/12