Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^{2} \sin \theta-x(\sin \theta \cos \theta+1)+\cos \theta=0$
$\left(0<\theta<45^{\circ}\right)$, and $\alpha<\beta$. Then $\sum_{n=0}^{\infty}\left(\alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}\right)$
is equal to :-
Correct Option: 1
$\mathrm{D}=(1+\sin \theta \cos \theta)^{2}-4 \sin \theta \cos \theta=(1-\sin \theta \cos \theta)^{2}$
$\Rightarrow$ roots are $\beta=\operatorname{cosec} \theta$ and $\alpha=\cos \theta$
$\Rightarrow \sum_{n=0}^{\infty}\left(\alpha^{n}+\left(-\frac{1}{\beta}\right)^{n}\right)=\sum_{n=0}^{\infty}(\cos \theta)^{n}+\sum_{n=0}^{n}(-\sin \theta)^{n}$
$=\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta}$
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