Question

If \( \cos(\theta - \alpha), \cos \theta \) and \( \cos(\theta + \alpha) \) are in harmonic progression, then \( 2 \tan^2 \theta = \)

• A. \( \tan^2 \frac{\alpha}{2} - 1 \)
• B. \( 1 + \tan^2 \frac{\alpha}{2} \)
• C. \( 1 + \cot^2 \frac{\alpha}{2} \)
• D. \( 1 - \cot^2 \frac{\alpha}{2} \)

Hint:

If \( a, b, c \) are in harmonic progression, then \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \). Use trigonometric identities to simplify the resulting equation and solve for \( \tan^2 \theta \). Remember the double and half angle formulas for cosine and sine.

Answer 1

The Correct Option is A

Since \( \cos(\theta - \alpha), \cos \theta, \cos(\theta + \alpha) \) are in harmonic progression, their reciprocals \( \frac{1}{\cos(\theta - \alpha)}, \frac{1}{\cos \theta}, \frac{1}{\cos(\theta + \alpha)} \) are in arithmetic progression.
Therefore, $$ \frac{2}{\cos \theta} = \frac{1}{\cos(\theta - \alpha)} + \frac{1}{\cos(\theta + \alpha)} $$ $$ \frac{2}{\cos \theta} = \frac{\cos(\theta + \alpha) + \cos(\theta - \alpha)}{\cos(\theta - \alpha) \cos(\theta + \alpha)} $$ Using the identities \( \cos(A + B) + \cos(A - B) = 2 \cos A \cos B \) and \( \cos(A - B) \cos(A + B) = \cos^2 A - \sin^2 B \): $$ \frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha} $$ $$ 2 (\cos^2 \theta - \sin^2 \alpha) = 2 \cos^2 \theta \cos \alpha $$ $$ \cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha $$ $$ \cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha $$ Using the identities \( 1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2} \) and \( \sin^2 \alpha = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} \): $$ \cos^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} $$ If \( \sin^2 \frac{\alpha}{2} \neq 0 \), we can divide by \( 2 \sin^2 \frac{\alpha}{2} \): $$ \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2} $$ Divide by \( \cos^2 \theta \): $$ 1 = 2 \frac{\cos^2 \frac{\alpha}{2}}{\cos^2 \theta} $$ $$ \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2} $$ We need to find \( 2 \tan^2 \theta \).
From \( \cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha \): $$ \cos^2 \theta = \frac{\sin^2 \alpha}{1 - \cos \alpha} = \frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}}{2 \sin^2 \frac{\alpha}{2}} = 2 \cos^2 \frac{\alpha}{2} $$ $$ \frac{1}{\sec^2 \theta} = 2 \frac{1}{1 + \tan^2 \frac{\alpha}{2}} $$ $$ 1 + \tan^2 \theta = \frac{1}{2} (1 + \tan^2 \frac{\alpha}{2}) $$ $$ 2 + 2 \tan^2 \theta = 1 + \tan^2 \frac{\alpha}{2} $$ $$ 2 \tan^2 \theta = \tan^2 \frac{\alpha}{2} - 1 $$