Let $A = \{\theta \in [0, 2\pi)$ : $\frac{1 + 2i \sin \theta}{1 - i \sin \theta}\}$ is purely imaginary.…

Let

$z=\frac{1+2i\sin \theta }{1-i\sin \theta }$

$\Rightarrow z=\frac{1+2i\sin \theta }{1-i\sin \theta }\times \frac{1+i\sin \theta }{1+i\sin \theta }$

$\Rightarrow z=\frac{1-2{\sin }^2\theta +3i\sin \theta }{1+{\sin }^2\theta }$

$\Rightarrow z=\left(\frac{1-2{\sin }^2\theta }{1+{\sin }^2\theta }\right)+i\left(\frac{3\sin \theta }{1+{\sin }^2\theta }\right)$

Since, $z$ is a purely imaginary number, so real part must be zero, hence

$\frac{1-2{\sin }^2\theta }{1+{\sin }^2\theta }=0$

$\Rightarrow 1-2{\sin }^2\theta =0$

$\Rightarrow \cos 2\theta =0$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{4},\frac{3\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},\frac{7\mathrm{\pi }}{4}$, for $\theta \in \left(0,2\mathrm{\pi }\right)$

$\therefore$Sum of all values $=\frac{16\mathrm{\pi }}{4}=4\mathrm{\pi }$