Q63 / 160 · AP EAMCET 2009 · Mathematics

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

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

Solution

$\begin{aligned} tan(x-y) \tan y=\frac{\sin (x-y) \sin y}{\cos (x-y) \cos y} \times \frac{2}{2} \\ =\frac{\cos (x-2 y)-\cos (x)}{\cos (x-2 y)+\cos (x)} \\ =\frac{1-\frac{\cos x}{\cos (x-2 y)}}{1+\frac{\cos (x)}{\cos (x-2 y)}} \\ =\frac{1-\lambda}{1+\lambda} \\ &\left(\text { Given, } \lambda=\frac{\cos x}{\cos (x-2 y)}\right) \end{aligned}$

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

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