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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

( x)/( (x-2 y))=λ (x-y) y is equal to

cos⁡xcos⁡(x−2y)=λ⇒tan⁡(x−y)tan⁡y\frac{\cos x}{\cos (x-2 y)}=\lambda \Rightarrow \tan (x-y) \tan y is equal to

Options

  1. A1+λ1−λ\frac{1+\lambda}{1-\lambda}
  2. B1−λ1+λ\frac{1-\lambda}{1+\lambda}
  3. Cλ1+λ\frac{\lambda}{1+\lambda}
  4. Dλ1−λ\frac{\lambda}{1-\lambda}

Correct answer

B. 1−λ1+λ\frac{1-\lambda}{1+\lambda}

Step-by-step solution

tan⁡(x−y)tan⁡y=sin⁡(x−y)sin⁡ycos⁡(x−y)cos⁡y×22=cos⁡(x−2y)−cos⁡(x)cos⁡(x−2y)+cos⁡(x)=1−cos⁡xcos⁡(x−2y)1+cos⁡(x)cos⁡(x−2y)=1−λ1+λ( Given, λ=cos⁡xcos⁡(x−2y))\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}
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