已知cos(a+b)=1/5,cos(a-b)=3/5,则tanatanb的值为谢谢了,如题

已知cos(a+b)cos(a-b)=1/3,则cos^2a-sin^2b的值是? Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2 已知锐角a、b满足cos a=4/5,tan(a-b)=-1/3,求cos b 已知cos(a+b)=-1/3,cos2a=-5/13,a,b均为锐角,求cos(a-b) 已知cos(a+b)=4/5,cos(a-b)=-4/5,3π/2 .已知Cos(a+b)=4/5,cos(a-b)= -4/5,3/2π 已知cos(a+b)cos(a-b)=1/3 ,求(cosa)^2-(cosb)^2 sin(a-b)*cos a-cos(a-b)*sin a=1/5,则cos 2b的值是 已知cos(a+b)cos(a-b)=1/4,则cos^2a+cos^2b=是（cosa)^2 已知cos(a+b)=1/5,cos(a-b)=3/5,则tana*tanb= 已知cos(a+b)=1/5、cos(a-b)=3/5,求tana .tanb的值? 已知cos(a+b)=1/5,cos(a-b)=3/5,求tanatanb的值 已知cos(a+b)=1/3,cos(a-b)=1/5,求tanatanb的值 已知cos(a+b)=1/3,cos(a-b)=1/5,求tana*tanb的值 已知cos（a+B)=1/3,cos(a-B)=1/5,求tana x tanB 已知cos(a +b) =1/3 ,cos (a-b) =1/5 ,求cosacosb 的值 已知cos（a+b）=2／3,cos(a-b)=1／5,则tanatanb=? 已知:5cos(a-b/2)+7cos(b/2)=0 求:tan(a/2)*tan((a-b)/2)已知 5cos(a-b/2)+7cos(b/2)=0,求tan(a/2)*tan[(a-b)/2].由5cos(a-b/2)+7cos(b/2)=0得5[cosacos(b/2)+sinasin(b/2)]+7cos(b/2)=05{[2cos^2(a/2)-1]cos(b/2)+sinasin(b/2)}+7cos(b/2)=010[cos^2(a/2)cos(b