若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?

问题描述:

若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?

(1+tanx)(1+tany)
=1+tanx+tany+tanxtany
=1+tan(x+y)*[1-tanxtany]+tanxtany
=2

x+y=π=4 所以tan(x+y)=tanπ/4=1=(tanx+tany)/(1-tanx*tany)
推出tanx+tany=1-tanxtany
(1+tanx)(1+tany)=1+tanxtany+tanx+tany=1+1-tanxtany+tanxtany=2