The correct option is A On n
△ = ∣∣
∣
∣∣111cos nxcos(n+1)xcos(n+2)xsin nxsin(n+1)xsin(n+2)x∣∣
∣
∣∣
Applying C1 → C1+C3−(2cos x)C2 △ = ∣∣
∣
∣∣2(1−cos x)110cos(n+1)xcos(n+2)x0sin(n+1)xsin(n+2)x∣∣
∣
∣∣ △ = 2(1−cos x)[cos(n+1)xsin(n+2)x−cos(n+2)xsin(n+1)x] △ = 2(1−cos x)[sin(n+2−n−1)x] = 2sin x(1−cos x)
i.e., △ is independent of n.