(sin x)^-1 is just csc x, so I'll write everything in terms of csc.
Add in the 'skipped' terms (csc 46)*(csc 47) + (csc 48)*(csc 49) + ... +(csc 134)*(csc 135). These are duplicate values of the terms given (but in reverse order), so the sum of the series with these extra terms is 2*csc(N).
Based on the rather convenient answer of N=1, I suspect this is a specific instance of an identity:
For any positive integer n:
2*csc(90/n) = sum{x=1 to n} csc(45+(x-1)*90/n)*csc(45+x*90/n)
Edited on June 24, 2017, 12:45 pm
An identity to prove
