Let a and b be the roots of the quadratic equation

Question:

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 :-

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

  2. $\frac{1}{1+\cos \theta}+\frac{1}{1-\sin \theta}$

  3. $\frac{1}{1-\cos \theta}-\frac{1}{1+\sin \theta}$

  4. $\frac{1}{1+\cos \theta}-\frac{1}{1-\sin \theta}$


Correct Option: 1

Solution:

$\mathrm{D}=(1+\sin \theta \cos \theta)^{2}-4 \sin \theta \cos \theta=(1-\sin \theta \cos \theta)^{2}$

$\Rightarrow$ 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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