The minimum and maximum values of $\sin^{2} (120 ° + θ) + \sin^{2} (120 ° - θ)$ are

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$\max = \frac{3}{2}, \min = \frac{1}{2}$

$\max = \frac{3}{2}, \min = \frac{1}{3}$

$\max = \frac{3}{2}, \min = 0$

$\max = \frac{1}{2}, \min = 0$

answer is A.

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Detailed Solution

$\sin^{2} (120 ° + θ) + \sin^{2} (120 ° - θ)$

$= \sin^{2} θ + \sin^{2} (120 + θ) + \sin^{2} (120 - θ) - \sin^{2} θ$

$\begin{array}{l} = \frac{3}{2} - \sin^{2} θ \\ 0 \leq \sin^{2} θ \leq 1 \\ \frac{1}{2} \leq \frac{3}{2} - \sin^{2} θ \leq \frac{3}{2} \end{array}$

Maximum value $= \frac{3}{2}$ , Minimum value $= \frac{1}{2}$

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