Let A={θ∈(0,2π):1+2isin⁡θ1−isin⁡θ is purely imaginary}.A = \left\{ \theta \in (0, 2\pi) : \frac{1+2i \sin \theta}{1-i \sin \theta} \text{ is purely imaginary} \right\}.…

Mathematics

Problem

Let A={θ∈(0,2π):1+2isin⁡θ1−isin⁡θ is purely imaginary}.A = \left\{ \theta \in (0, 2\pi) : \frac{1+2i \sin \theta}{1-i \sin \theta} \text{ is purely imaginary} \right\}. Then, the sum of the elements in AA is (2023 Main, 8 April II) (a) 2π2\pi (b) 3π3\pi (b) π\pi (d) 4π4\pi

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