The resulting matrix won't be very sparse. Anywhere your Hscos has a 0 (including the implicitly stored 0's) the result will contain a nonfinite, non-zero value.
Hscos = speye(5)
y = sparse(1)./Hscos
fullY = full(y); % for comparison
whos Hscos y fullY
You're consuming more memory storing y (every element of which is non-zero) as a sparse matrix than you would be if you stored it as a full matrix!
One potential approach would be to extract the non-zero elements from your sparse matrix Hscos, divide by the elements in the corresponding location in the other matrix, and treat anything with a 0 in Hscos as giving a 0 in the result. I don't know if that's the best approach for the underlying problem you're trying to solve, but if those 0's are actually important then you're going to need to operate on the full versions of your matrices.
