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Let $A = \{θ \in (- \frac{π}{2}, π) : \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ} is purely imaginary\}$ . Then the sum of the elements in A .is

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detailed solution

Correct option is B

Let $z = \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ}$

$= \frac{3 + 2 i \sin θ}{1 - 2 i \sin θ} \times \frac{1 + 2 i \sin θ}{1 + 2 i \sin θ} = \frac{(3 - 4 \sin^{2} θ) + i (8 \sin θ)}{1 + 4 \sin^{2} θ}$

Now, $(z) = 0$ [ $∵ z$ is purely imaginary]

$\begin{array}{l} \Rightarrow \frac{3 - 4 \sin^{2} θ}{1 + 4 \sin^{2} θ} = 0 \Rightarrow \sin^{2} θ = \frac{3}{4} \\ \Rightarrow \sin^{2} θ = \sin^{2} \frac{π}{3} ∴ θ = n π \pm \frac{π}{3}, n \in Z \end{array}$

$θ = - \frac{π}{3}, \frac{π}{3}, \frac{2 π}{3}$ $[∵ θ \in (\frac{- π}{2}, π)]$

Hence, sum of elements of $f (A) = \frac{- π}{3} + \frac{π}{3} + \frac{2 π}{3} = \frac{2 π}{3}$

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Let A=θ∈−π2,π:3+2isin⁡θ1−2isin⁡θ is purely imaginary . Then the sum of the elements in A .is