已知cos（a+b）=0.2 ,cos（a-b）=0.6 求tana*tanb已知cos（a+b）=0.2 ,cos（a-b）=0.6求tana*tanb

已知： cos(A+B)=0.2,cos(A-B)=0.6.
求：tanAtanB.
∵cos(A+B)=0.2,cos(A-B)=0.6
将两式展开
cosAcosB-sinAsinB=0.2…①
cosAcosB+sinAsinB=0.6…②
联立两式可得sinAsinB=0.2,cosAcosB=0.4
∴tanAtanB=(sinAsinB)/( cosAcosB)=0.5.

0.5
cos（a+b）=cosacosb-sinasinb=0.2````````1式
cos（a-b）=cosacosb+sinasinb=0.6````````2式
1式加2式等于cosacosb=0.4
1式减2式等于sinasinb=0.2
tana*tanb =0.2/0.4=0.5
给分吧！

由cos（a+b）=0.2 得 cosacosb-sinasinb=0.2 ①由cos（a-b）=0.6 得 cosacosb+sinasinb=0.6 ②①+②得 2cosacosb=0.8,即cosacosb=0.4②-①得 2sinasinb=0.4,即sinasinb=0.2②÷①可得 tanatanb=sinasinb/cosacosb=0....

cos（a+b）=0.2 ，cos（a-b）=0.6
∴cosacosb-sinasinb=0.2 ①
cosacosb+sinasinb=0.6 ②
①+②得2cosacosb=0.8，cosacosb=0.4
②-①得2sinasinb=0.4，sinasinb=0.2
∴tanatanb=sinasinb/cosacosb=0.2/0.4=0.5