Question

If $ \cos A = m \cos B $ \text{ and } $ \cot \left( \frac{A+B}{2} \right) = \lambda \tan \left( \frac{B-A}{2} \right), $ \text{ then } $ \lambda \text{ is equal to} $

• A. \( \frac{m}{m-1} \)
• B. \( \frac{m+1}{m} \)
• C. \( \frac{m+1}{m-1} \)
• D. None of these

Hint:

When dealing with trigonometric equations, use known identities like the tangent and cotangent identities to express angles in terms of each other and simplify the problem.

Answer 1

The Correct Option is C

Given: \[ \cos A = m \cos B \quad \text{and} \quad \cot \left( \frac{A+B}{2} \right) = \lambda \tan \left( \frac{B-A}{2} \right) \] From the first equation, we have: \[ \cos A = m \cos B \] Now, using the trigonometric identity for cotangent and tangent: \[ \cot \left( \frac{A+B}{2} \right) = \frac{1}{\tan \left( \frac{A+B}{2} \right)} \quad \text{and} \quad \tan \left( \frac{B-A}{2} \right) = \frac{\sin \left( \frac{B-A}{2} \right)}{\cos \left( \frac{B-A}{2} \right)} \] Equating the two sides from the second equation: \[ \cot \left( \frac{A+B}{2} \right) = \lambda \tan \left( \frac{B-A}{2} \right) \] After solving the trigonometric expressions and simplifying, we arrive at: \[ \lambda = \frac{m+1}{m-1} \]