已知2tanA=3tanB,求证tan(A+B)=sin2B/5-cos2B 已知a ,b为锐角,cosa=4/5,tan(a-b)=1/3 ,求cosb
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/31 08:02:21
xRN@~۲Җ$5'/H@#I!&%?A/F/ ̖T8lgfn":*!z4~?]'^բ^vxny1W(@h6OD2sh<G/ӶxWL_d;dZ0=Kctp
C`5NPVUifhHfVx1_9QueC@ލ@6@<" 16Vu[rNh~hU²'}+Y0,%
SS/sf1ǝVXmVF|&k{fB+064&SfIcfe͆WxSY
已知tan(A-B)/tanA+sin^2C/sin^2A=1,求证:tanA×tanB=tan^C求证tanA×tanB=tan^2C
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B.
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B
已知:tan(a-b)=sin2a/(5-coa2b),求证2tana=3tanb
已知2tanA=3tanB.求证:tan(A-B)=sin2B/5-cos2B
已知tan(A-B)=sin2B/(5-cos2B) 求证:2tanA=3tanB
已知tan(A-B)/tanA+sin^2C/sin^2A=1,求证:tanA*tanB=tan^2C
已知A、B是锐角,求证(tan(π+A)+tan(-B))/(1/tan(3π-A)+tan(π/2-B))=tanA*tanB
已知A、B是锐角,求证(tan(π+A)+tan(-B))/(1/tan(3π-A)+tan(π/2-B))=tanA*tanB
必修二,三角函数已知2tanA=3tanB求证tan(A-B)=sin2B/5-cos2B
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B(帮忙用word发下吧)
已知:(1+tanA)(1+tanB)=2,求证:tan(A+B)=1
已知2tanA=tanB,求证:tan(A-B)=sin2B/(5-cos2B)
已知sin(2a+b)=3sinb,设tana=x,tanb=y求证:tan﹙a+b)=2tana
已知tana+tanb=2,tan(a+b)=4.则tana*tanb
2tana=3tanb,求证tan(a-b)=sin2b/5-cos2b.
已知tan(a+b)=3,tana乘tanb=3/2,则tana+tanb=(有解题过程)tanb=3分之2
求证:化简[tan(A+B)-tanA-tanB]÷[tanA·tan(A+B)]=tanB.RT