解:因为cos(α+β)=cosαcosβ-sinαsinβ =1/5,①cos(α-β)=cosαcosβ+sinαsinβ =2/5,②①+②可得:2cosαcosβ=3/5,则cosαcosβ=3/10,②-①可得:2sinαsinβ=1/5,则sinαsinβ=1/10,所以tanαtanβ=sinαsinβ/cosαcosβ=(1/10)/(3/10)=1/3.
cosacosb-sinasinb=1/5cosacosb+sinasinb=2/5两个相除可得(cosacosb-sinasinb)/(cosacosb+sinasinb)=1/2分子分母都除以cosacosb可得(1-tanatanb)/(1+tanatanb)=1/2设tanatanb=K(1-K0/(1+K)=1/21+K=2(1-K)K=1/3tanatanb=1/3