若a+b=3π/4,求(1-tana)(1-tanb)

(1-tana)(1-tanb)
=1-tana-tanb+tanatanb
tan(a+b)=(tana+tanb)/(1-tanatanb)
=tan(3π/4)=-1
即tana+tanb=tanatanb-1
所以
(1-tana)(1-tanb)
=1-tana-tanb+tanatanb
=1+tanatanb-(tana+tanb)
=1+tanatanb-(tanatanb-1)
=2