Question

\( \frac{\cos x}{\cos(x - 2y)} = \lambda \Rightarrow \tan(x - y)\tan y \) is equal to

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

Hint:

Use $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$.

Answer 1

The Correct Option is B

Step 1: Expansion
Write $\tan(x-y) \tan y$ as $\frac{\sin(x-y) \sin y}{\cos(x-y) \cos y}$.
Step 2: Product-to-Sum
$\frac{2 \sin(x-y) \sin y}{2 \cos(x-y) \cos y} = \frac{\cos((x-y)-y) - \cos((x-y)+y)}{\cos((x-y)-y) + \cos((x-y)+y)} = \frac{\cos(x-2y) - \cos x}{\cos(x-2y) + \cos x}$.
Step 3: Substitute given ratio
Divide numerator and denominator by $\cos(x-2y)$ to get $\frac{1 - \frac{\cos x}{\cos(x-2y)}}{1 + \frac{\cos x}{\cos(x-2y)}}$.
Step 4: Result
$= \frac{1-\lambda}{1+\lambda}$.
Final Answer: (B)