haiyouyidao

(a+1)(b+1)=2
ab+a+b+1=2
a+b=1-ab

tan(arctana)=a,tan(arctanb)=b
所以tan(arctana+arctanb)=[tan(arctana)+tan(arctanb)]/[1-tan(arctana)tan(arctanb)]
=(a+b)/(1-ab)
=1

-π/2<arctana<π/2
-π/2<arctanb<π/2
-π<arctana+arctanb<π

若ab都小于0，或都大于0，都可以满足(a+1)(b+1)=0
所以arctana和arctanb可正可负
所以arctana+arctanb=π/4或-3π/4