已知sin(a+β)=1/2,sin(a-β)=1/3,则tan²βtan(a+β)分之(tan(a+β)-tana-tanβ）=?
网友回答
(tan(a+β)-tana-tanβ)/tan²βtan(a+β)
=[tan(a+β)-t(a+β)(1-tana*tanβ)]/tan²βtan(a+β)
=tan(a+β)[1-(1-tana*tanβ)]/tan²βtan(a+β)
=tana*tanβ/tan²β=tana/tanβ
=2sina*cosβ/2cosa*sinβ
=[(sinacosβ+cosasinβ)+(sinacosβ-cosasinβ)]/[(sinacosβ+cosasinβ)-(sinacosβ-cosasinβ)]
=[sin(a+β)+sin(a-β)]/[sin(a+β)-sin(a-β)]
=5======以下答案可供参考======
供参考答案1:
原式=5