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

Substitute the values of cosα and cosβ in terms of the tangents of their half angles. cosα=(1−tan²(α/2))/(1+tan²(α/2)) and cosβ=(1−tan²(β/2))/(1+tan²(β/2)) Let tan²(α/2)=a and tan²(β/2)=b Substituting the above values in the expression, 2((1−b)/(1+b)−(1−a)/(1+a)) + ((1−a)/(1+a))((1−b)/(1+b))=1 ⇒2((1−b+a−ab)−(1−a+b−ab))/(1+a+b+ab) + (1−a−b+ab)/(1+a+b+ab)=1 ⇒2(a−b)/(1+a+b+ab) + (1−a−b+ab)/(1+a+b+ab)=1 ⇒2(a−b)+1−a−b+ab=1+a+b+ab ⇒4(a−b)+1−a−b+ab=1+a+b+ab ⇒4a−4b=2b+2a ⇒2a=6b ⇒a=3b ⇒tan²(α/2)=3tan²(β/2) ⇒tan²(α/2)−3tan²(β/2)=0 ⇒(tan(α/2)+√3tan(β/2))(tan(α/2)−√3tan(β/2))=0 ∴tan(α/2)+√3tan(β/2)=0 or tan(α/2)−√3tan(β/2)=0 Hence options C and D are correct.