prove that 1+ 3cosA-4cos^3/ 1- cosA = (1+2cosA)^2
1 + 3cosa - 4(cosa)^3/(1-cosa) = (1+2cosa)^2
(1 + 3cosa)(1 - cosa) - 4(cosa)^3 = (1 + 4cosa + 4(cosa)^2) (1 - cosa)
1 + 2cosa - 3(cosa)^2 - 4(cosa)^3 = 1 + 4cosa + 4(cosa)^2 - cosa - 4(cosa)^2 - 4(cosa)^3
1 + 2cosa - 3(cosa)^2 - 4(cosa)^3 = 1 + 3cosa - 4(cosa)^3
2cosa - 3(cosa)^2 = 3cosa
-3(cosa)^2 = cosa
-3cosa = 1
cosa = -1/3
Not always true; not ture for all values of A.
If this is supposed to end up being always true, the problem should result in an answer like cosa = cosa or 1 = 1. If that's what should have happened, you might want to re-write the problem with special attention on parentheses.
Or I might have made a mistake, in which case someone can feel free to point the error out.