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(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula,$x = \frac{(\sin

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$\frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$

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(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula,$x = \frac{(\sin 	heta \cos 	heta + 1) \pm \sqrt{(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta}}{2 \sin 	heta}$.Simplifying the discriminant: $(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta = (\sin 	heta \cos 	heta - 1)^2$.Thus,$x = \frac{\sin 	heta \cos 	heta + 1 \pm (\sin 	heta \cos 	heta - 1)}{2 \sin 	heta}$.This gives $x_1 = \frac{2 \sin 	heta \cos 	heta}{2 \sin 	heta} = \cos 	heta$ and $x_2 = \frac{2}{2 \sin 	heta} = \csc 	heta$.Since $0  \sin 	heta$,so $\cos 	heta The sum is $S = \sum_{n=0}^\infty \alpha^n + \sum_{n=0}^\infty (-\frac{1}{\beta})^n = \frac{1}{1 - \alpha} + \frac{1}{1 + 1/\beta} = \frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$.

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$ where $0 < \theta < 45^\circ$,and $\alpha < \beta$. Then $\sum_{n=0}^\infty (\alpha^n + \frac{(-1)^n}{\beta^n})$ is equal to

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