sinθcosθsin2θsin(θ + 2π3)cos(θ + 2π3)sin(2θ + 4π3)sin(θ - 2π3)cos(θ - 2π3)sin(2θ - 4π3)

R2→R2+R3 =sinθcosθsin2θ sinθ + 2π3+sinθ-2π3 cosθ + 2π3+cosθ-2π3 sin2θ + 4π3+sinθ-4π3 sinθ - 2π3cosθ - 2π3 sin2θ - 4π3 =sinθcosθsin2θ 2sinθ cos2π3 2cosθ cos2π3 2sin2θ cos4π3 sinθ-2π3cosθ-2π3sin2θ-4π3 =sinθcosθsin2θ -sinθ-cosθ-sin2θ sinθ-2π3cosθ-2π3sin2θ-4π3 R1→R1+R2 =000 -sinθ-cosθ-sin2θ sinθ-2π3cosθ-2π3sin2θ-4π3 =0

sinθcosθsin2θsin(θ + 2π3)cos(θ + 2π3)sin(2θ + 4π3)sin(θ - 2π3)cos(θ - 2π3)sin(2θ - 4π3)

R2→R2+R3 =sinθcosθsin2θ sinθ + 2π3+sinθ-2π3 cosθ + 2π3+cosθ-2π3 sin2θ + 4π3+sinθ-4π3 sinθ - 2π3cosθ - 2π3 sin2θ - 4π3 =sinθcosθsin2θ 2sinθ cos2π3 2cosθ cos2π3 2sin2θ cos4π3 sinθ-2π3cosθ-2π3sin2θ-4π3 =sinθcosθsin2θ -sinθ-cosθ-sin2θ sinθ-2π3cosθ-2π3sin2θ-4π3 R1→R1+R2 =000 -sinθ-cosθ-sin2θ sinθ-2π3cosθ-2π3sin2θ-4π3 =0

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$|\begin{matrix} \sin θ & \cos θ & \sin 2 θ \\ \sin (θ + \frac{2 π}{3}) & \cos (θ + \frac{2 π}{3}) & \sin (2 θ + \frac{4 π}{3}) \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix}|$

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$cosθ + \sin^{2} θ$

$\sin θ$

answer is D.

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Detailed Solution

$R_{2} \to R_{2} + R_{3} = |\begin{matrix} sinθ & cosθ & \sin 2 θ \\ \sin (θ + \frac{2 π}{3}) + \sin (θ - \frac{2 π}{3}) & \cos (θ + \frac{2 π}{3}) + \cos (θ - \frac{2 π}{3}) & \sin (2 θ + \frac{4 π}{3}) + \sin (θ - \frac{4 π}{3}) \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix}| = |\begin{matrix} sinθ & cosθ & \sin 2 θ \\ 2 \sin θ \cos (\frac{2 π}{3}) & 2 cosθ \cos \frac{2 π}{3} & 2 \sin 2 θ \cos (\frac{4 π}{3}) \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix}|$ $= |\begin{matrix} sinθ & cosθ & \sin 2 θ \\ - sinθ & - cosθ & - \sin 2 θ \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix}| R_{1} \to R_{1} + R_{2} = |\begin{matrix} 0 & 0 & 0 \\ - sinθ & - cosθ & - \sin 2 θ \\ \sin (θ - \frac{2 π}{3}) & \cos (θ - \frac{2 π}{3}) & \sin (2 θ - \frac{4 π}{3}) \end{matrix}| = 0$