Verified Solution • 2025–26 Session moderate

QJEEAdv17Ma9fcd: JEE Advanced 2017 Mathematics - Class 11 JEEAdv17Ma9fcd Solution: Let and be non zero real numbers such that . Then which of the…

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

Multiple Choice Options

  • A 3tan⁡(α2)−tan⁡(β2)=2\sqrt 3 \tan \left( {{\alpha \over 2}} \right) - \tan \left( {{\beta \over 2}} \right) = 2
  • B tan⁡(α2)−3tan⁡(β2)=0\tan \left( {{\alpha \over 2}} \right) - \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0
  • C tan⁡(α2)+3tan⁡(β2)=0\tan \left( {{\alpha \over 2}} \right) + \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0
  • D 3tan⁡(α2)+tan⁡(β2)=2\sqrt 3 \tan \left( {{\alpha \over 2}} \right) + \tan \left( {{\beta \over 2}} \right) = 2

Step-by-Step Verified Solution

2(cos⁡β−cos⁡α)+cos⁡αcos⁡β=12(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1

or 4(cos⁡β−cos⁡α)+2cos⁡αcos⁡β=24(\cos \beta - \cos \alpha ) + 2\cos \alpha \cos \beta = 2

⇒1−cos⁡α+cos⁡β−cos⁡αcos⁡β\Rightarrow 1 - \cos \alpha + \cos \beta - \cos \alpha \cos \beta

=3+3cos⁡α−3cos⁡β−3cos⁡αcos⁡β= 3 + 3\cos \alpha - 3\cos \beta - 3\cos \alpha \cos \beta

⇒(1−cos⁡α)(1+cos⁡β)\Rightarrow (1 - \cos \alpha )(1 + \cos \beta )

=3(1+cos⁡α)(1−cos⁡β)= 3(1 + \cos \alpha )(1 - \cos \beta )

⇒(1−cos⁡α)(1+cos⁡α)=3(1−cos⁡β)1+cos⁡β\Rightarrow {{(1 - \cos \alpha )} \over {(1 + \cos \alpha )}} = {{3(1 - \cos \beta )} \over {1 + \cos \beta }}

⇒tan⁡2α2=3tan⁡2β2\Rightarrow {\tan ^2}{\alpha \over 2} = 3{\tan ^2}{\beta \over 2}

∴\therefore tan⁡α2±3tan⁡β2=0\tan {\alpha \over 2} \pm \sqrt 3 \tan {\beta \over 2} = 0

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