Let α,β, and γ be such that 0<α<β<γ<2π. For any x∈R if cosx+α+cosx+β+cosx+γ=0 then tanγ-α=

cosx+α+cosx+β+cosx+γ=0 cosx(cosα+cosβ+cosγ)-sinx(sinα+sinβ+sinγ)=0 cosα+cosβ+cosγ=0⇒cosα+cosγ=-cosβ----(i) sinα+sinβ+sinγ=0⇒sinα+sinγ=-sinβ----(ii) (i)2+(ii)2⇒ cos2α+cos2γ+2cosα cosγ+sin2α+sin2γ+2sinαsinγ=cos2β+sin2β 2+2(cosα cosγ+sinαsinγ)=1 cos(α-γ)=-12⇒α-γ=2π3 tan(γ-α)=tan-2π3=3

Let α,β, and γ be such that 0<α<β<γ<2π. For any x∈R if cosx+α+cosx+β+cosx+γ=0 then tanγ-α=

cosx+α+cosx+β+cosx+γ=0 cosx(cosα+cosβ+cosγ)-sinx(sinα+sinβ+sinγ)=0 cosα+cosβ+cosγ=0⇒cosα+cosγ=-cosβ----(i) sinα+sinβ+sinγ=0⇒sinα+sinγ=-sinβ----(ii) (i)2+(ii)2⇒ cos2α+cos2γ+2cosα cosγ+sin2α+sin2γ+2sinαsinγ=cos2β+sin2β 2+2(cosα cosγ+sinαsinγ)=1 cos(α-γ)=-12⇒α-γ=2π3 tan(γ-α)=tan-2π3=3

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$Let α, β, and γ be such that 0 < α < β < γ < 2 π . For any x \in R if \cos (x + α) + \cos (x + β) + \cos (x + γ) = 0 then \tan (γ - α) =$

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$- \sqrt{3}$

$\sqrt{3}$

answer is D.

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Detailed Solution

$\cos (x + α) + \cos (x + β) + \cos (x + γ) = 0 \cos x (\cos α + \cos β + \cos γ) - \sin x (\sin α + \sin β + \sin γ) = 0 \cos α + \cos β + \cos γ = 0 \Rightarrow \cos α + \cos γ = - \cos β - - - - (i) \sin α + \sin β + \sin γ = 0 \Rightarrow \sin α + \sin γ = - \sin β - - - - (i i) {(i)}^{2} + {(i i)}^{2} \Rightarrow \cos^{2} α + \cos^{2} γ + 2 \cos α \cos γ + \sin^{2} α + \sin^{2} γ + 2 \sin α \sin γ = \cos^{2} β + \sin^{2} β 2 + 2 (\cos α \cos γ + \sin α \sin γ) = 1 \cos (α - γ) = - \frac{1}{2} \Rightarrow α - γ = \frac{2 π}{3} \tan (γ - α) = \tan (- \frac{2 π}{3}) = \sqrt{3}$