若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?

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若x=π/16,y=3π/16,则(1+tanx)(1+tany)=?若x=π/16,y=3π/16,则(1+tanx)(1+tany)=?若x=π/16,y=3π/16,则(1+tanx)(1+ta

若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?
若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?

若x=π/16,y=3π/16 ,则(1+tanx)(1+tany)=?
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

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