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Let $\mathrm{A}=\left\{\theta \in(0,2 \pi): \frac{1+2 \mathrm{i} \sin \theta}{1-\mathrm{i} \sin \theta}\right.$ is purely imaginary $\}$. Then the sum of the elements in A is [JEE (Main) 08 April 2023 SII]
(A) $3 \pi$
(B) $2 \pi$
(C) $4 \pi$
(D) $\pi$

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Solution

$\mathrm{Z}=\frac{1+2 \mathrm{i} \sin \theta}{1-\mathrm{i} \sin \theta}$ is purely imaginary So, $\operatorname{Re}(\mathrm{z})=0 \Rightarrow \frac{1-2 \sin ^{2} \theta}{1+\sin ^{2} \theta}=0$
$\Rightarrow \sin ^{2} \theta=\frac{1}{2} \therefore \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$, for $\theta \in(0,2 \pi) \therefore$ Sum of all values $=\frac{16 \pi}{4}=4 \pi$

[UPLIFTJEE]

Answer

C

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