Question

Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation \(x^2\sin\theta - x(\sin\theta\cos\theta + 1) + \cos\theta = 0\) \((0 \lt \theta \lt 45^{\circ})\), and \(\alpha \lt \beta\). Then \(\sum_{n=0}^{\infty}\left(\alpha^n + \frac{(-1)^n}{\beta^n}\right)\) 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. \(\frac{1}{1+\cos\theta} - \frac{1}{1-\sin\theta}\)

Hint:

Factor the quadratic as (x sin(theta) - 1)(x - cos(theta)); identify which root is smaller and sum the two geometric series.

Answer 1

The Correct Option is B

\(x^2\sin\theta - x(\sin\theta\cos\theta + 1) + \cos\theta = (x\sin\theta - 1)(x - \cos\theta)\), so the roots are \(\frac{1}{\sin\theta}\) and \(\cos\theta\).
For \(0 \lt \theta \lt 45^\circ\): \(\cos\theta \lt 1 \lt \frac{1}{\sin\theta}\), so \(\alpha = \cos\theta\) and \(\beta = \frac{1}{\sin\theta}\).
\(\sum_{n=0}^{\infty}\alpha^n = \frac{1}{1-\cos\theta}\) (since \(0 \lt \cos\theta \lt 1\)).
\(\frac{(-1)^n}{\beta^n} = (-\sin\theta)^n\), so \(\sum_{n=0}^{\infty}(-\sin\theta)^n = \frac{1}{1+\sin\theta}\).
Total = \(\frac{1}{1-\cos\theta} + \frac{1}{1+\sin\theta}\).
Answer: (2) \(\frac{1}{1-\cos\theta} + \frac{1}{1+\sin\theta}\)