Question

If $ \frac{\cos x}{\cos(x - 2y)} = \lambda \quad \text{then} \quad \tan(x - y) \tan y = $

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

Hint:

When simplifying trigonometric equations, remember to use standard trigonometric identities such as \( \cos(x - y) = \cos x \cos y + \sin x \sin y \) and \( \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \).

Answer 1

The Correct Option is B

We are given: \[ \frac{\cos x}{\cos(x - 2y)} = \lambda \]
Step 1: Express the equation in terms of trigonometric identities
Using the trigonometric identity for \( \cos(x - 2y) \), we can expand: \[ \cos(x - 2y) = \cos x \cos 2y + \sin x \sin 2y \] Substitute this in the given equation: \[ \frac{\cos x}{\cos x \cos 2y + \sin x \sin 2y} = \lambda \]
Step 2: Simplify the equation
\[ \frac{1}{\cos 2y + \tan x \sin 2y} = \lambda \] Now, multiply both sides by \( \cos 2y + \tan x \sin 2y \): \[ 1 = \lambda(\cos 2y + \tan x \sin 2y) \] Now, we want to calculate \( \tan(x - y) \tan y \).
Step 3: Using the formula for \( \tan(x - y) \)
We can use the identity for \( \tan(x - y) \): \[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \] Thus, \[ \tan(x - y) \tan y = \frac{(\tan x - \tan y) \cdot \tan y}{1 + \tan x \tan y} \]
Step 4: Final Simplification
Simplifying this expression using the previously derived equation, we get the result: \[ \tan(x - y) \tan y = \frac{1 - \lambda}{1 + \lambda} \] Thus, the correct answer is (B).