if t=tan[theta/2],
tan[theta]=2t/(1-t^2)
cot[theta/2]=1/t
.'. you get {[2t/(1-t^2)] - t}/{1/t + [2t/(1-t^2)]}
= {[2t-t(1-t^2)]/(1-t^2)}/{[1-t^2+2t^2]/[t(1-t^2)]}
= {[2t-t+t^3]/(1-t^2)}/{[1+t^2]/[t(1-t^2)]}
= {[t(1+t^2]/(1-t^2)}/{[1+t^2]/t(1-t^2)]}
from this you can cancel some of the top and bottom parts out, and you end up with:
t/[1/t] = t^2
p.s. sorry, i dunno how to use latex