(sin x + 2 cos x)/(3 sin x + 4 cos x)

Put, Numerator = A (Denominator) + B [d/dx (Denominator)]
∴ sin x+ 2 cos x = A(3 sin x + 4 cos x) + B [d/dx (3 sin x + 4 cos x)]
= A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x)
∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x
Equating the coefficients of sin x and cos x on both the sides, we get
3A – 4B = 1 …… (1)
and 4A + 3B = 2 …… (2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
9A – 12B = 3
16A + 12B = 8
On adding, we get
