Question

If \( \theta + \phi = \alpha \) and \( \tan \theta = k \tan \phi \) (where \( k > 1 \)), then the value of \( \sin(\theta - \phi) \) is

• A. \( k \tan \phi \)
• B. \( \sin \alpha \)
• C. \( \left( \frac{k-1}{k+1} \right) \sin \alpha \)
• D. \( k \cos \phi \)

Hint:

Componendo-dividendo is useful when a ratio is given. For tangent, use \(\tan A \pm \tan B = \frac{\sin(A \pm B)}{\cos A \cos B}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Given \(\theta + \phi = \alpha\) and \(\tan \theta = k \tan \phi\) with \(k > 1\). We need \(\sin(\theta - \phi)\).

Step 2: Key Formula or Approach:
Use componendo-dividendo on the given tangent ratio.

Step 3: Detailed Explanation:
\(\tan \theta = k \tan \phi \implies \frac{\tan \theta}{\tan \phi} = k\).
Apply componendo-dividendo: \[ \frac{\tan \theta + \tan \phi}{\tan \theta - \tan \phi} = \frac{k+1}{k-1}. \] Using \(\tan \theta + \tan \phi = \frac{\sin(\theta + \phi)}{\cos \theta \cos \phi}\) and \(\tan \theta - \tan \phi = \frac{\sin(\theta - \phi)}{\cos \theta \cos \phi}\), we get: \[ \frac{\sin(\theta + \phi)}{\sin(\theta - \phi)} = \frac{k+1}{k-1}. \] Since \(\theta + \phi = \alpha\), \(\sin(\theta + \phi) = \sin \alpha\). Therefore, \[ \frac{\sin \alpha}{\sin(\theta - \phi)} = \frac{k+1}{k-1} \implies \sin(\theta - \phi) = \left( \frac{k-1}{k+1} \right) \sin \alpha. \]

Step 4: Final Answer:
Option (C) is correct.