Trigonometric Identity

How to prove that [ sin^(2)(x) + 2cos(x) - 1 ] / [ sin^(2)(x) + 3cos(x) -3 ] = 1 / [ 1 - sec(x) ]. I tried to use sin^(2)(x) = 1 - cos^(2)(x) and I can factor out a cos(x) in the numerator of LHS but not sure what I can do with the denominator.

Please see attached question for what I tried to do... Maybe there is another way of simplifying the LHS?

okraina,
It's easy. Set all the Sin^2 terms to (1-Cos^2)^2 , The Sec term is 1/Cos. All the terms will be Cos factors and the equation balance, thereby proving the equality correct. Use a computer to do all the messy algebra. Ratch.