cos(θ−α),cosθ,cos(θ+α) are in H.P then cosθ.sec(α2)=

cos(θ−α), cosθ, cos(θ+α) are in H.P 1cos(θ−α)1cosθ1cos(θ+α) are in H.P 2cosθ=1cos(θ−α)+1cos(θ+α) 2cosθ=2cosθcosαcos(θ−α)cos(θ+α) cos2θcosα=cos2θ−sin2α ⇒cos2θ(1−cosα)=sin2α cos2θ=1+cosα⇒cos2θ=2cos2(α2) cos2θsec2α2=2⇒cosθsecα2=±2

cos(θ−α),cosθ,cos(θ+α) are in H.P then cosθ.sec(α2)=

cos(θ−α), cosθ, cos(θ+α) are in H.P 1cos(θ−α)1cosθ1cos(θ+α) are in H.P 2cosθ=1cos(θ−α)+1cos(θ+α) 2cosθ=2cosθcosαcos(θ−α)cos(θ+α) cos2θcosα=cos2θ−sin2α ⇒cos2θ(1−cosα)=sin2α cos2θ=1+cosα⇒cos2θ=2cos2(α2) cos2θsec2α2=2⇒cosθsecα2=±2

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$\cos (θ - α), \cos θ, \cos (θ + α)$ are in H.P then $\cos θ . \sec (\frac{α}{2}) =$

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$- \sqrt{2}$

$\sqrt{2}$

$\pm \sqrt{2}$

answer is C.

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Detailed Solution

$\cos (θ - α), \cos θ, \cos (θ + α) a r e i n H . P \frac{1}{\cos (θ - α)} \frac{1}{\cos θ} \frac{1}{\cos (θ + α)} a r e i n H . P \frac{2}{\cos θ} = \frac{1}{\cos (θ - α)} + \frac{1}{\cos (θ + α)} \frac{2}{\cos θ} = \frac{2 \cos θ \cos α}{\cos (θ - α) \cos (θ + α)} \cos^{2} θ \cos α = \cos^{2} θ - \sin^{2} α \Rightarrow \cos^{2} θ (1 - \cos α) = \sin^{2} α \cos^{2} θ = 1 + \cos α \Rightarrow \cos^{2} θ = 2 \cos^{2} (\frac{α}{2}) \cos^{2} θ \sec^{2} \frac{α}{2} = 2 \Rightarrow \cos θ \sec \frac{α}{2} = \pm \sqrt{2}$