Question

If \( \theta \) is an acute angle, \( x = \sum_{n=0}^{\infty} \cos^{2n} \theta \), \( y = \sum_{n=0}^{\infty} \sin^{2n} \theta \), then \( \frac{1}{x^2} + \frac{1}{y^2} = \)

• A. \( 1 - \frac{2}{xy} \)
• B. \( 1 - \frac{1}{xy} \)
• C. \( 1 + \frac{1}{xy} \)
• D. \( 1 + \frac{2}{xy} \)

Hint:

For these variables, \( x+y = xy \) is a very useful property. From this, it follows that \( 1/x + 1/y = 1 \). Recognizing these patterns quickly simplifies complex algebraic manipulations.

Answer 1

The Correct Option is A

Concept: The sum of an infinite geometric progression (G.P.) is \( S_{\infty} = \frac{a}{1-r} \) where \( |r| < 1 \).
• \( x \) and \( y \) are infinite G.P. sums.
• Trigonometric identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).

Step 1: Evaluating \( x \) and \( y \).
Since \( \theta \) is acute, \( 0 < \cos^2\theta < 1 \) and \( 0 < \sin^2\theta < 1 \). \[ x = 1 + \cos^2\theta + \cos^4\theta + \dots = \frac{1}{1 - \cos^2\theta} = \frac{1}{\sin^2\theta} \implies \sin^2\theta = \frac{1}{x} \] \[ y = 1 + \sin^2\theta + \sin^4\theta + \dots = \frac{1}{1 - \sin^2\theta} = \frac{1}{\cos^2\theta} \implies \cos^2\theta = \frac{1}{y} \]

Step 2: Finding the relationship between \( x \) and \( y \).
From \( \sin^2\theta + \cos^2\theta = 1 \): \[ \frac{1}{x} + \frac{1}{y} = 1 \]

Step 3: Calculating \( \frac{1}{x^2} + \frac{1}{y^2} \).
Square the identity from
Step 2: \[ \left( \frac{1}{x} + \frac{1}{y} \right)^2 = 1^2 \] \[ \frac{1}{x^2} + \frac{1}{y^2} + \frac{2}{xy} = 1 \] \[ \frac{1}{x^2} + \frac{1}{y^2} = 1 - \frac{2}{xy} \]