∵:sinB=(√10)/10,
∴:cosB=√[1-(sinB)^2]=(3√10)/10
tanB=sinB/cosB=[(√10)/10]/[(3√10)/10]=1/3
又:tanA=1/7
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=(1/7+1/3)/[1-(1/7)(1/3)]
=1/2
∵:sinB=(√10)/10,
∴:cosB=√[1-(sinB)^2]=(3√10)/10
tanB=sinB/cosB=[(√10)/10]/[(3√10)/10]=1/3
又:tanA=1/7
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=(1/7+1/3)/[1-(1/7)(1/3)]
=1/2