Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^{2} \sin \theta-x(\sin \theta \cos…

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

Solution

$x^{2} \sin \theta-x(\sin \theta \cdot \cos \theta+1)+\cos \theta=0$. $x^{2} \sin \theta-x \sin \theta \cdot \cos \theta-x+\cos \theta=0$ $x \sin \theta(x-\cos \theta)-1(x-\cos \theta)=0$ $(x-\cos \theta)(x \sin \theta-1)=0$ $\therefore x=\cos \theta, \operatorname{cosec} \theta, \theta \in\left(0,45^{\circ}\right)$ $\therefore \alpha=\cos \theta, \beta=\operatorname{cosec} \theta$ $\sum_{n=0}^{\infty} \alpha^{n}=1+\cos \theta+\cos ^{2} \theta+\ldots \infty=\frac{1}{1-\cos \theta}$ $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{\beta^{n}}=1-\frac{1}{\operatorname{cosec} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta}-\frac{1}{\operatorname{cosec}^{3} \theta}+\ldots \infty$ $=1-\sin \theta+\sin ^{2} \theta-\sin ^{3} \theta+\ldots \infty$ $=\frac{1}{1+\sin \theta}$ $\therefore \quad \sum_{n=0}^{\infty}\left(\alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}\right)=\sum_{n=0}^{\infty} \alpha^{n}+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{\beta^{n}}$ $=\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta} .$

Asked in: JEE Main 2019 (11 Jan Shift 2)

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