Let A = \ θ ∈ (0, 2π): (1 + 2isinθ)/(1 - isinθ)ispurelyimaginary\...

Let

$z = \dfrac{1 + 2isin\theta}{1 - isin\theta}$

$⇒ z = \dfrac{1 + 2isin\theta}{1 - isin\theta} \times \dfrac{1 + isin\theta}{1 + isin\theta}$

$⇒ z = \dfrac{1 - 2\sin^{2}\theta + 3isin\theta}{1 + \sin^{2}\theta}$

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

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

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

$⇒ 1 - 2\sin^{2}\theta = 0$

$⇒ \cos 2\theta = 0$

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

$∴$Sum of all values $= \dfrac{16\pi}{4} = 4\pi$