Question

已知sin(a+β)=2/3,sin(a-β)=2/5,求

求(tana-tanβ)/(tana+tanβ)的值

Answer 1

(tanα-tanβ)/(tanα+tanβ)

=(sinαcosβ-sinβcosα)/(sinαcosβ+sinβcosα)
=sin(α-β)/sin(α+β)
=3/5

Answer 2

显然a和β的三角函数值没有0
sin(a+β)=sinacosβ+cosasinβ=2/3
两边同时除以cosacosβ，得：tana+tanβ=2/3cosacosβ
sin(a-β)=sinacosβ-cosasinβ=2/5
两边同时除以cosacosβ，得：tana-tanβ=2/5cosacosβ
两式相除，得(tana-tanβ)/(tana+tanβ)=3/5

Answer 3

sin(a+β)=2/3 得到sinacosβ+sinβcosa=2/3
sin(a-β)=2/5 得到sinacosβ-sinβcosa=2/5
解出sinacosβ=8/15 sinβcosa=2/15
得到tana/tanβ=sinacosβ/sinβcosa=4
(tana-tanβ)/(tana+tanβ）=（tana/tanβ-1）/（1+tana/tanβ）=（4-1）/（4+1）=3/5

Answer 4

(tana-tanβ)/(tana+tanβ)＝﹙sinαcosβ－sinβcosα﹚/﹙sinαcosβ＋sinβcosα﹚

＝sin(a-β)/sin(a+β)＝3/5

Answer 5

(tana-tanβ)/(tana+tanβ)
=[(sinacosb-sinbcosa)/cosacosb]/[(sinacosb+sinbcosa)/cosacosb]
=(sinacosb-sinbcosa)/(sinacosb+sinbcosa)
=sin(a-β)/sin(a+β)
=3/5