求y=arctanx+arctan[(1-x)/(1+x)]的值
tany=[tan(arctanx)+tanarctan(1-x)/(1+x)]/[1-tan(arctanx)tanarctan(1-x)/(1+x)]
=[x+(1-x)/(1+x)]/[1-x(1-x)/(1+x)]=[x(1+x)+(1-x)]/[(1+x)-x(1-x)]=(x²+1)/(1+x²)=1
故y=π/4.