Question

已知cos（α+β）=1∕5，cos(α-β）=3∕5，求tanαtanβ的值

Answer 1

cos(a+b)=1/5
cosacosb-sinasinb=1/5(1)
cos(a-b)=3/5
cosacosb+sinasinb=3/5（2）
(1)+(2)
2cosacosb=4/5
cosacosb=2/5(3)
(2)-(1)
2sinasinb=2/5
sinasinb=1/5(4)
(4)/(3)
sinasinb/cosacosb=1/2
tanatanb=1/2

Answer 2

cos(a+B)=cosa*cosB-sina*sinB=1/5,
cos(a-B)=~~~~~~~~+~~~~~~~=3/5,
联立求解，得(1)cosa*cosB=2/5,(2)sina*sinB=1/5,
tana*tanB=(2)/(1)=1/2