Answer: (3)

Solution

$x^2\sin\theta - x(\sin\theta\cos\theta + 1) + \cos\theta = 0$

$x\sin\theta(x - \cos\theta) - 1(x - \cos\theta) = 0$

$(x - \cos\theta)(x\sin\theta - 1) = 0$

$\therefore x = \cos\theta, \csc\theta$, $\theta \in (0, 45°)$

$\therefore \alpha = \cos\theta, \beta = \csc\theta$

$\sum_{n=0}^{\infty}\alpha^n = 1 + \cos\theta + \cos^2\theta + \ldots = \frac{1}{1-\cos\theta}$

$\sum_{n=0}^{\infty}\frac{(-1)^n}{\beta^n} = 1 - \frac{1}{\csc\theta} + \frac{1}{\csc^2\theta} - \ldots = \frac{1}{1+\sin\theta}$

$\therefore \sum_{n=0}^{\infty}\left(\alpha^n + \frac{(-1)^n}{\beta^n}\right) = \frac{1}{1-\cos\theta} + \frac{1}{1+\sin\theta}$