Question

已知0<β<α<π/2,cos(α+β)=-3/5，sin(α-β)=5/13

（1）求tanαtanβ
（2）求tanα/tanβ

Answer 1

由0<β<α<π/2，cos(α+β)=-3/5，sin(α-β)=5/13

得 0<α+β<π ，-π/2<α-β<π/2

所以 sin(α+β)=4/5，cos(α-β)=12/13

sinαsinβ = -1/2 [cos(α+β)-cos(α-β)] = 99/130

cosαcosβ = 1/2 [cos(α+β)+cos(α-β)] =21/130

sinαcosβ = 1/2 [sin(α+β)+sin(α-β)] =77/130

cosαsinβ = 1/2 [sin(α+β)-sin(α-β)] =27/130

（1）tanαtanβ = sinαsinβ/cosαcosβ = 33/7

（2）tanα/tanβ =sinαcosβ/cosαsinβ = 77/27