1.Let A={θ∈(−π2,π):3+2isinθ1−2isinθ is purely imaginary}A = \left\{\theta \in \left(-\dfrac{\pi}{2}, \pi\right) : \dfrac{3 + 2i\sin\theta}{1 - 2i\sin\theta} \text{ is purely imaginary}\right\}A={θ∈(−2π,π):1−2isinθ3+2isinθ is purely imaginary}. Then the sum of the elements in AAA is:a.5π6\dfrac{5\pi}{6}65πb.π\piπc.2π3\dfrac{2\pi}{3}32πd.3π4\dfrac{3\pi}{4}43πLogin to continueOnly logged in users canattempt or see the solution.
