若（cosA-cosB+1/2)^2+(sinA-sinB-1/4)^2=0,求tanAtanB的值

tanAtanB=39/231或tanAtanB=231/39.

由(cosA-cosB+1/2)^2+(sinA-sinB-1/4)^2=0得,
cosA-cosB+1/2=0,sinA-sinB-1/4=0
cosA-cosB＝-1/2,sinA-sinB＝1/4
将上面两式平方得
(cosA)^2+(cosB)^2-2cosAcosB=1/4
(sinA)^2+(sinB)^2-2sinAsinB=1/16
两式相加得
cosAcosB+sinAsinB=27/32
cos(A-B)=27/32
再将cosA-cosB=-1/2,sinA-sinB=1/4两式相乘得
sinAcosA-sinAcosB-cosAsinB+sinBcosB=-1/8
sin2A+sin2B-2sin(A+B)=-1/4，和差化积得
sin(A+B)cos(A-B)-sin(A+B)=-1/8，将cos(A-B)=27/32代入得
sin(A+B)(27/32)-sin(A+B)=-1/8
sin(A+B)=4/5,利用(sin(A+B)）^2+(sin(A+B)）^2=1,求得
cos(A+B)=3/5或cos(A+B)=-3/5
如果cos(A+B)=3/5，则cos(A+B)+cos(A-B)=3/5+27/32=231/160
即cosAcosB=231/320,
cos(A+B)-cos(A-B)=3/5-27/32=-39/160
即sinAsinB=39/320,故得
tanAtanB=(sinAsinB)/(cosAcosB)=39/231.
如果cos(A+B)=-3/5，则cos(A+B)+cos(A-B)=-3/5+27/32=39/160
即cosAcosB=39/320,
cos(A+B)-cos(A-B)=-3/5-27/32=-231/160
即sinAsinB=231/320,故得
tanAtanB=(sinAsinB)/(c