Let , be the roots of x^2 - x(+1)+=0 (0

Question

Let $\alpha,\beta$ be the roots of $x^2\sin	heta - x(\sin	heta\cos	heta+1)+\cos	heta=0$ ($0<	heta<45^{\circ}$), $\alpha<\beta$. Then $\displaystyle\sum_{n=0}^{\infty}\left[\alpha^n+\frac{(-1)^n}{\beta^n}ight]$ is equal to:

Accepted Answer

Factoring: $(x-\cos\theta)(x\sin\theta-1)=0 \Rightarrow x=\cos\theta,\ \csc\theta$; so $\alpha=\cos\theta$, $\beta=\csc\theta$.

$$\sum\alpha^n = \frac{1}{1-\cos\theta}, \qquad \sum\frac{(-1)^n}{\beta^n} = \sum(-\sin\theta)^n = \frac{1}{1+\sin\theta}.$$

$$\text{Total} = \frac{1}{1-\cos\theta}+\frac{1}{1+\sin\theta}.$$

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