I don't think A^2+B^=C^2 method is the right way to go about the problem. Kw loading will stay the same, but changing the kvar will cause the kva to change.
Page 4 of (
https://www.eaton.com/ecm/groups/pub...a02607001e.pdf) has a good visual of the change.
The OP's method seems the quickest to answer the problem:
1. Solve for kw. (pf x KVAinitial =kw) --- 0.75 x 750000 = 562500 Kw
2. Find the angle for .95 pf. (cos-1 [Pfdesired]= angle) --- cos-1 (0.95) = 18.19 degrees
3. Solve for Kvar. (tan[angle] x kw = kvar) --- tan(18.19) x 562500 = 184831 Kvar
For fun you can then use A^2+B^=C^2 to solve for the new Kvar = 592088 Kvar