已知f(x)=根号(1-x)/1+x，若a∈（π/2,π），则f(cosa)+f(-cosa)可化简为

∵ a∈（π/2,π） ∴a/2∈(π/4,π/2) ∴sin(a/2)>0,cos(a/2)>0

f(cosa)+f(-cosa)=√[(1-cosa)/(1+cosa)]+√[(1+cosa)/(1-cosa)]...①

利用公式cosx=2[cos(x/2)]^2-1=1-2[sin(x/2)]^2,可化简

1-cosa=1-{1-2[sin(a/2)]^2}=2[sin(a/2)]^2
1+cosa=1+{2[cos(a/2)]^2-1}=2[cos(a/2)]^2 代入①式，得

f(cosa)+f(-cosa)=√{2[sin(a/2)]^2}/{2[cos(a/2)]^2}+√{2[cos(a/2)]^2}/{2[sin(a/2)]^2}

=[sin(a/2)]/[cos(a/2)]+[cos(a/2)]/[sin(a/2)]

={[sin(a/2)]^2+[cos(a/2)]^2}/[cos(a/2)sin(a/2)]

=1/[(1/2)sina]=2/sina