求y=arctanx+arctan(1-x)/(1+x)的值
x∈（-无穷,-1）∪（-1,+无穷）
tany=tan(arctanx+arctan(1-x)/(1+x))=1
∵x≠-1,(1-x)/(1+x)≠-1,arctanx∈(-π/2,-π/4)∪(-π/4,π/2）,arctan(1-x)/(1+x)∈(-π/2,-π/4)∪(-π/4,π/2) ,
又∵tany=1,
∴y=π/4或-3π/4.
最后一步不太懂.为什么会分两类呢
当x＞-1时,y=π/4；当x＜-1时,y=3π/4.