Question bank/JEE Main/Mathematics

Let $A = \left\{ \theta \in \left( - \frac{\pi}{2},\pi \right):\frac{3 + 2isin\theta}{1 - 2isin\theta}\ ispurely\ imaginary\ \right\}.$ Then
the sum of the elements in A is

A$\frac{2\pi}{3}$B$\pi$C$\frac{5\pi}{6}$D$\frac{3\pi}{4}$

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Let  $z = \frac{3 + 2isin\theta}{1 - 2isin\theta}$
θ$\in \left( - \frac{\pi}{2},\pi \right)$
For z to be purely imaginary
$\frac{3 + 2{isin}\theta}{1 - 2{isin}\theta} + \frac{3 - 2{isin}\theta}{1 + 2{isin}\theta} = 0$
$\Rightarrow$
$\frac{3 + 6isin\theta + 2isin\theta - 4\sin^{2}\theta + 3 - 6isin\theta - 2isin\theta - 4\sin^{2}\theta}{1 + 4\sin^{2}\theta}$
$\Rightarrow$ $3 - 4\sin^{2}\theta = 0 \Rightarrow$
$\sin^{2}\theta = \frac{3}{4}$
$\theta = n\pi \pm \frac{\pi}{3}$
Given $\theta \in \left( - \frac{\pi}{2},\pi \right)$
Hence $\theta = - \frac{\pi}{3},\frac{\pi}{3},\frac{2\pi}{3}$
sum of all possible value of $\theta = \frac{2\pi}{3}$

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx