Question

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

• A. $\frac{1}{1 - \cos \theta} + \frac{1}{1+ \sin \theta} $
• B. $\frac{1}{1 + \cos \theta} + \frac{1}{1 + \sin \theta} $
• C. $\frac{1}{1 - \cos \theta} - \frac{1}{1+ \sin \theta} $
• D. $\frac{1}{1 + \cos \theta} - \frac{1}{1 - \sin \theta} $

Answer 1

Answer: (A)

$D =\left(1+\sin\theta \cos\theta \right)^{2} - 4 \sin\theta \cos\theta =\left(1-\sin\theta \cos\theta \right)^{2}$
$ \Rightarrow $ roots are $\beta =\cos ec \theta$ and $ \alpha=\cos\theta $
$ \Rightarrow \sum^{\infty}_{n=0} \left(\alpha^{n} + \left(- \frac{1}{\beta}\right)^{n} \right) = \sum^{\infty}_{n=0} \left(\cos\theta\right)^{n} + \sum^{n}_{n=0} \left(-\sin\theta\right)^{n} $
$ = \frac{1}{1- \cos\theta} + \frac{1}{1+\sin\theta}$