MathematicsMedium53×since 2002Q3163Let A={θ∈(0,2π):1+2isinθ1−isinθA=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.A={θ∈(0,2π):1−isinθ1+2isinθ is purely imaginary }\}}. Then the sum of the elements in A\mathrm{A}A is :A3π3 \pi3πBπ\piπC2π2 \pi2πD4π4 \pi4πCheck answerSkip