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

## The correct options are C tan(α2)+√3tan(β2)=0 D tan(α2)−√3tan(β2)=0Substitute the values of cosα and cosβ in terms of the tangents of their half angles.cosα=1−tan2α21+tan2α2 and cosβ=1−tan2β21+tan2β2Let tan2α2=a and tan2β2=bSubstituting the above values in the expression,2(1−b1+b−1−a1+a)+(1−a1+a)(1−b1+b)=1⇒2(1−b+a−ab−1+a−b+ab1+a+b+ab)+1−a−b+ab1+a+b+ab=1⇒4(a−b)+1−a−b+ab=1+a+b+ab⇒4a−4b=2b+2a⇒2a=6b⇒a=3b⇒tan2α2=3tan2β2⇒tan2α2−3tan2β2=0⇒(tanα2+√3tanβ2)(tanα2−√3tanβ2)=0∴tanα2+√3tanβ2=0 or tanα2−√3tanβ2=0Hence options C and D are correct.

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