已知x+y=pai/4,则(1+tanx)(1+tany)=

问题描述:

已知x+y=pai/4,则(1+tanx)(1+tany)=

tan(x+y)=tanpai/4=1
(tanx+tany)/(1-tanxtany)=1
tanx+tany=1-tanxtany
tanx+tany+tanxtany=1
原式=1+tanx+tany+tanxtany
=1+1
=2