Как вычислить: (2sin2a-3cos2a)/(4sin2a+5cos2a), если tga=3 ?

sin(a) = 3cos(a)
sin^2(a) = 9cos^2(a)
1 - cos^2(a) = 9cos^2(a)
10cos^2(a) = 1
cos^2(a) = 0.1
... = (4sin(a)cos(a) - 3(1 - 2sin^2(a))) / (8sin(a)cos(a) + 5(1 - 2sin^2(a))) = (4cos(a)3cos(a) - 3 + 6*9cos^2(a)) / (8cos(a)3cos(a) + 5 - 10*9cos^2(a)) = (12cos^2(a) + 54cos^2(a) - 3) / (24cos^2(a) - 90cos^2(a) + 5) = (66cos^2(a) - 3) / (-66cos^2(a) + 5) = (6.6 - 3) / (5 - 6.6) = 3.6 / (-1.6) = -2.25