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JEE Advanced 2017Paper 2
Exam
JEE Advanced
Year
2017
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Advanced 2017 · Paper 2 · Mathematics · Trigonometric Equations

Let α and β be non zero real numbers such that 2(cos β - cos α) +...

Let α\alpha and β\beta be non zero real numbers such that 2(cosβ−cosα)+cosαcosβ=12\left(cos \beta - cos \alpha\right) + cos \alpha cos \beta = 1. Then which of the following is/are true?

Options

  1. A\tan \left(\dfrac{α}{2}\right) - \sqrt{3}\tan \left(\dfrac{β}{2}\right) = 0
  2. B3tan⁡(α2)−tan⁡(β2)=0\sqrt{3}\tan \left(\dfrac{\alpha}{2}\right) - \tan \left(\dfrac{\beta}{2}\right) = 0
  3. C\tan \left(\dfrac{α}{2}\right) + \sqrt{3}\tan \left(\dfrac{β}{2}\right) = 0
  4. D3tan⁡(α2)+tan⁡(β2)=0\sqrt{3}\tan \left(\dfrac{\alpha}{2}\right) + \tan \left(\dfrac{\beta}{2}\right) = 0

Correct answer

A. \tan \left(\dfrac{α}{2}\right) - \sqrt{3}\tan \left(\dfrac{β}{2}\right) = 0

Step-by-step solution

2\left(cos β - cos α\right) + cos αcos β - 1 = 0 .....(i)

Use cos β = \dfrac{1 - \tan^{2} \dfrac{β}{2}}{1 + \tan^{2} \dfrac{β}{2}}andcos α = \dfrac{1 - \tan^{2} \dfrac{α}{2}}{1 + \tan^{2} \dfrac{α}{2}} in (i)

We get \tan^{2} \left(\dfrac{α}{2}\right) = 3\tan^{2} \left(\dfrac{β}{2}\right) \to \tan \left(\dfrac{α}{2}\right) + \sqrt{3}tan \dfrac{β}{2} = 0ortan \left(\dfrac{α}{2}\right) - \sqrt{3}\tan \left(\dfrac{β}{2}\right) = 0

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