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

JEE Main · Mathematics · Trigonometric Ratios & Identities

If (x-y), x, (x+y) are three distinct numbers which are in harmonic...

If cos⁡(x−y),cos⁡x,cos⁡(x+y)\cos (x-y), \cos x, \cos (x+y) are three distinct numbers which are in harmonic progression and cos⁡x≠cos⁡y\cos x \neq \cos y, then 1+cos⁡y1+\cos y is equal to

Options

  1. Acos⁡2x\cos ^2 x
  2. B−cos⁡2x-\cos ^2 x
  3. Ccos⁡2x−1\cos ^2 x-1
  4. Dcos⁡2x−2\cos ^2 x-2

Correct answer

A. cos⁡2x\cos ^2 x

Step-by-step solution

cos⁡(x−y),cos⁡x,cos⁡(x+y)\cos (x-y), \cos x, \cos (x+y) are in HP. Then, cos⁡x=2cos⁡(x−y)cos⁡(x+y)cos⁡(x+y)+cos⁡(x−y)cos⁡x=cos⁡2x+cos⁡2y2cos⁡x⋅cos⁡ycos⁡x=2cos⁡2x+2cos⁡2y−22cos⁡x⋅cos⁡ycos⁡2x⋅cos⁡y=cos⁡2x+cos⁡2y−1cos⁡2x(cos⁡y−1)=(cos⁡2y−1)cos⁡2x(1−cos⁡y)=(1−cos⁡2y)cos⁡2x(2sin⁡2y2)=sin⁡2ycos⁡2x(2sin⁡2y2)=(2sin⁡y2⋅cos⁡y2)2cos⁡2x(2sin⁡2y2)=4sin⁡2y2⋅cos⁡2y2cos⁡2x=(2cos⁡2y2−1)+1cos⁡y+1=cos⁡2x\cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y)+\cos (x-y)}\cos x=\frac{\cos 2 x+\cos 2 y}{2 \cos x \cdot \cos y}\cos x=\frac{2 \cos ^2 x+2 \cos ^2 y-2}{2 \cos x \cdot \cos y}\cos ^2 x \cdot \cos y=\cos ^2 x+\cos ^2 y-1\cos ^2 x(\cos y-1)=\left(\cos ^2 y-1\right)\cos ^2 x(1-\cos y)=\left(1-\cos ^2 y\right)\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\sin ^2 y\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\left(2 \sin \frac{y}{2} \cdot \cos \frac{y}{2}\right)^2\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=4 \sin ^2 \frac{y}{2} \cdot \cos ^2 \frac{y}{2}\cos ^2 x=\left(2 \cos ^2 \frac{y}{2}-1\right)+1\cos y+1=\cos ^2 x or 1+cos⁡y=cos⁡2x\quad 1+\cos y=\cos ^2 x
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