Q109 / 180 · COMEDK 2024 (Evening Shift) · Mathematics

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

  • A
    $\dfrac{\lambda}{1-\lambda}$
  • B
    $\dfrac{1+\lambda}{1-\lambda}$
  • Correct answer
    C
    $\dfrac{1-\lambda}{1+\lambda}$
  • D
    $\dfrac{\lambda}{1+\lambda}$

Solution

Given $\dfrac{\cos x}{\cos (x-2y)} = \lambda$.

Applying componendo and dividendo:

$\dfrac{\cos (x-2y) + \cos x}{\cos (x-2y) - \cos x} = \dfrac{1+\lambda}{1-\lambda}$

Using the sum-to-product formulas $\cos A + \cos B = 2 \cos \left( \dfrac{A+B}{2} \right) \cos \left( \dfrac{A-B}{2} \right)$ and $\cos A - \cos B = -2 \sin \left( \dfrac{A+B}{2} \right) \sin \left( \dfrac{A-B}{2} \right)$:

$\dfrac{2 \cos \left( \dfrac{x-2y+x}{2} \right) \cos \left( \dfrac{x-2y-x}{2} \right)}{-2 \sin \left( \dfrac{x-2y+x}{2} \right) \sin \left( \dfrac{x-2y-x}{2} \right)} = \dfrac{1+\lambda}{1-\lambda}$

$\dfrac{2 \cos (x-y) \cos (-y)}{-2 \sin (x-y) \sin (-y)} = \dfrac{1+\lambda}{1-\lambda}$

Since $\cos (-y) = \cos y$ and $\sin (-y) = -\sin y$, the expression simplifies to:

$\dfrac{\cos (x-y) \cos y}{\sin (x-y) \sin y} = \dfrac{1+\lambda}{1-\lambda}$

$\cot (x-y) \cot y = \dfrac{1+\lambda}{1-\lambda}$

Taking the reciprocal of both sides:

$\tan (x-y) \tan y = \dfrac{1-\lambda}{1+\lambda}$

Answer: $\dfrac{1-\lambda}{1+\lambda}$

Hence, the correct answer is option: 'C'.

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