1.Let A={θ∈(0,2π):1+2isinθ1−isinθ is purely imaginary}A = \left\{\theta \in (0, 2\pi) : \dfrac{1 + 2i\sin\theta}{1 - i\sin\theta} \text{ is purely imaginary}\right\}A={θ∈(0,2π):1−isinθ1+2isinθ is purely imaginary}. Then the sum of the elements in AAA is:a.4π4\pi4πb.3π3\pi3πc.π\piπd.2π2\pi2πLogin to continueOnly logged in users canattempt or see the solution.
