已获知cos(A+B)=1/5,cos(A-B)=3/5,求tanAtanB

问题描述:

已获知cos(A+B)=1/5,cos(A-B)=3/5,求tanAtanB

cos(A+B)=cosAcosB-sinAsinB=1/5
cos(A-B)=cosAcosB+sinAsinB=3/5
所以cosAcosB=2/5,sinAsinB=1/5
那么tanAtanB=sinAsinB/cosAcosB=(1/5)/(2/5)=1/2