已知tan(a+β)=7,tana*tanβ=2/3则cos(a-β)=
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已知tan(a+β)=7,tana*tanβ=2/3则cos(a-β)=
![已知tan(a+β)=7,tana*tanβ=2/3则cos(a-β)=](/uploads/image/z/8628680-56-0.jpg?t=%E5%B7%B2%E7%9F%A5tan%28a%2B%CE%B2%29%3D7%2Ctana%2Atan%CE%B2%3D2%2F3%E5%88%99cos%28a-%CE%B2%29%3D)
tan(a+β)=7
(tana+tanβ)/(1-tanatanβ) = 7
3(tana+tanβ) = 7
3sin(a+β)/cosacosβ = 7
cosacosβ= 3sin(a+β)/7
= 3(7/5√2)/7
= 3/(5√2) = 3√2/10
tanatanβ = 2/3
sinasinβ/cosacosβ = 2/3
sinasinβ = (2/3)cosacosβ = (2/3)(3√2/10)= √2/5
cos(a-β) =cosacosβ+ sinasinβ
= 3√2/10 + √2/5
= √2/2
(tana+tanβ)/(1-tanatanβ) = 7
3(tana+tanβ) = 7
3sin(a+β)/cosacosβ = 7
cosacosβ= 3sin(a+β)/7
= 3(7/5√2)/7
= 3/(5√2) = 3√2/10
tanatanβ = 2/3
sinasinβ/cosacosβ = 2/3
sinasinβ = (2/3)cosacosβ = (2/3)(3√2/10)= √2/5
cos(a-β) =cosacosβ+ sinasinβ
= 3√2/10 + √2/5
= √2/2
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