Multiplying by cos(x/2), both numerator and denominator,{1 - tan(x/2)}/{1 + tan(x/2)} = {cos(x/2) - sin(x/2)}/{cos(x/2) + sin(x/2)}

2) Then rationalizing, by multiplying with {cos(x/2) - sin(x/2)}{1 - tan(x/2)}/{1 + tan(x/2)} = {cos(x/2) - sin(x/2}²/{cos²(x/2) - sin²(x/2)}= {cos²(x/2) + sin²(x/2) - 2sin(x/2)*cos(x/2)}/{cos(x)} [Expanding the numeration using (a-b)² identity and by double angle identity denominator = cos(x)]This further simplifies to: {1 - sin(x)}/cos(x) [Since by identity, cos²A + sin²A = 1 and by multiple angle identity, sin(2A) = 2sin(A)cos(A)]

3) Thus the question simplifies to: {1 - sin(x)}²/[{cos(x)}*(π - 2x)³]

4) Let π - 2x = t; So, x --> π/2, t --> 0.Also x = π/2 - t/2==> cos(x) = cos(π/2 - t/2) = sin(t/2) = 2sin(t/4)cos(t/4)

As well, 1 - sin(x) = 1 - sin(π/2 - t/2) = 1 - cos(t/2) = 2sin²(t/4) [By multiple angle identity, {1 - cos(2x) = 2sin²x]==> {1 - sin(x)}² = 4sin⁴(t/4)

5) As of the above, the question transforms to:Lim (t --> 0) [{4sin⁴(t/4)}/{2sin(t/4)cos(t/4)(t³)}]

This simplifies to: Lim (t --> 0) [2sin³(t/4)/{(t³)*cos(t/4)}]

= 2*Lim (t --> 0) sin³(t/4)/t³ [As t --> 0, cos(t/4) = cos(0) = 1]

= (2/64)*Lim (t --> 0) [{sin(t/4)/(t/4)}³]

= (2/64)*1 = 1/32