Question

高数问题求解

Answer 1

令arctan[(1+x)/(1-x)]=α，arctanx=β
则，tanα=(1+x)/(1-x)，tanβ=x
那么，tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)
=[(1+x)/(1-x)-x]/[1+(1+x)x/(1-x)]
=[(1+x)-x(1-x)]/[(1-x)+(1+x)x]
=(1+x-x+x²)/(1-x+x+x²)
=1
所以，α-β=arctan1=π/4

Answer 2

公式：
tan(x+y)=(tanx+tany)/(1-tanxtany),tan(π/4)=1,tan(arctanx)=x.
(1-tanx)/(1+tanx)=(tanπ/4-tanx/)(1+tanπ/4tanx)=tan(π/4-x)
(1-x)/(1+x)=(tanπ/4-arctanx)(1+tanπ/4arctanx)=tan(π/4-arctanx)
arctan[(1-x)/(1+x)]=arctan[tan(π/4-arctanx)]=π/4-arctanx