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Verified Answer :

I am assuming you want to prove: 
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x). 

If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get: 
LHS = csc(x)/[1 - cos(x)] 
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]} 
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares 
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x). 

Then, since csc(x) = 1/sin(x): 
LHS = {csc(x)[1 + cos(x)]}/sin^2(x) 
= {[1 + cos(x)]/sin(x)}/sin^2(x) 
= [1 + cos(x)]/sin^3(x) 
= RHS. 

I hope this helps!

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