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Question

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

A
3tan(α2)tan(β2)=0
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B
tan(α2)3tan(β2)=0
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C

tan(α2)+3tan(β2)=0
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D
3tan(α2)+tan(β2)=0
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Solution

The correct options are B tan(α2)−√3tan(β2)=0 C tan(α2)+√3tan(β2)=0 We have, 2(cosβ−cosα)+cosαcosβ=1 Or 4(cosβ−cosα)+2cosαcosβ=2 ⇒1−cosα+cosβ−cosαcosβ =3+3cosα−3cosβ−3cosααcosβ ⇒(1−cosα)(1+cosβ)=3(1+cosα)(1−cosβ) ⇒(1−cosα)(1+cosα)=3(1−cosβ)(1+cosβ) ⇒tan2α2=3tan2β2 ∴ tanα2±√3tanβ2=0

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