1-(1-t/1+t)^2/1+(1-t/1+t))^2 怎样答案算出来是2t/1+t^2 求高手帮帮忙~~~~(>_

{1-[(1-t)/(1+t)]^2}/{1+[(1-t)/(1+t)]^2 }
={1-[(1-t)/(1+t)]^2}/{1+[(1-t)/(1+t)]^2}
={1-[(1-t)/(1+t)]^2}/{1+(1-t)^2/(1+t)^2}
={1-[(1-t)/(1+t)]^2}/{[(1+t)^2+(1-t)^2]/(1+t)^2}
={1-[(1-t)/(1+t)]^2}/{[1+t^2+2t+1+t^2-2t]/(1+t)^2}
={1-[(1-t)^2/(1+t)^2}/{[2+2t^2]/(1+t)^2}
={(1+t)^2/(1+t)^2-(1-t)^2/(1+t)^2}/{[2+2t^2]/(1+t)^2}
={[(1+t)^2-(1-t)^2]/(1+t)^2}/{[2+2t^2]/(1+t)^2}
={[1+t^2+2t-1-t^2+2t]/(1+t)^2}/{[2+2t^2]/(1+t)^2}
={4t/(1+t)^2}/{[2+2t^2]/(1+t)^2}
=4t/(1+t)^2*(1+t)^2/(2+2t^2)
=4t/(2+2t^2)
=2t/(1+t^2)