Little messy.
Forgive me Krizalid, but as you know, we can't use LaTex, it might take you a while to decode what I did.
the question is:
Prove: (sin(x) - cos(x))/sqrt(sin(2x)) = [sin(x) - cos(x) - cot(2x)*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]
Consider RHS:
[sin(x) - cos(x) - cot(2x)*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]
= [sin(x) - cos(x) - {cos(2x)/sin(2x)}*sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]
= [sin(x) - cos(x) - cos(2x)/sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]
= [{sin(x)*sqrt(sin(2x)) - cos(x)*sqrt(sin(2x)) - cos(2x)}/sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))] .......combine the top fractions
= [sin(x)*sqrt(sin(2x)) - cos(x)*sqrt(sin(2x)) - cos(2x)]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))] .......combine the top and bottom into one fraction
Let's leave that one there
Now, consider the LHS:
(sin(x) - cos(x))/sqrt(sin(2x))
= (sin(x) - cos(x))/sqrt(sin(2x)) * [sin(x) + cos(x) + sqrt(sin(2x))]/[sin(x) + cos(x) + sqrt(sin(2x))]
= [sin^2(x) + sin(x)cos(x) + sin(x)*sqrt(sin2x)) - sin(x)cos(x) - cos^2(x) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]
= [-(cos^2(x) - sin^2(x)) + sin(x)*sqrt(sin2x)) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]
= [-cos(2x) + sin(x)*sqrt(sin2x)) - cos(x)*sqrt(sin(2x))]/[{sin(x) + cos(x) + sqrt(sin(2x))}*sqrt(sin(2x))]
now we see that both sides "simplify" to the same thing. thus LHS = RHS
QED