tan[arctan(1+x)]=1+x
tan[arctan(1-x)]=1-x
所以tan[arctan(1+x)+arctan(1-x)]=[(1+x)+(1-x)]/[1-(1+x)(1-x)]
=tan1/4π=1
所以2/x^2=1
x^2=2
arccos(x/2)=arccos(±√2/2)
arccos(√2/2)=1/4π
arccos(-√2/2)=3/4π
选C
tan[arctan(1+x)]=1+x
tan[arctan(1-x)]=1-x
所以tan[arctan(1+x)+arctan(1-x)]=[(1+x)+(1-x)]/[1-(1+x)(1-x)]
=tan1/4π=1
所以2/x^2=1
x^2=2
arccos(x/2)=arccos(±√2/2)
arccos(√2/2)=1/4π
arccos(-√2/2)=3/4π
选C