JEE Math Practice Question
Let $$A=\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1-i \sin \theta}\right.$$ is purely imaginary $$\}$$. Then the sum of the elements in $$\mathrm{A}$$ is :
- A.$$3 \pi$$
- B.$$\pi$$
- C.$$2 \pi$$
- D.$$4 \pi$$
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