1cos(θ-α) cosθ cos(θ+α) → HP
So
1cos(θ-α) 1cosθ 1cos(θ+α) → AP
Using formula of Arithmetic Mean,
2cosθ = 1cos(θ-α) + 1cos(θ+α)
2cosθ = cos(θ+α) + cos(θ-α)cos(θ+α) . cos(θ-α)
2cosθ = 2cosθcosαcos2θ - sin2α
1cosθ = 1cosθcosαcos2θ - sin2α
cos2θ - sin2α = cos2θ. cosα
cos2θ (1- cosα) = sin2α
cos2θ (1- cosα) = 1-cos2α
cos2θ (1- cosα) = (1+cosα)(1-cosα)
cos2θ = 1+cosα
cosθ + cosα = 1
Since -1<cosθ<1, so from the above equation,
either
cosθ=1 and cosα=0...
OR
cosθ=0 and cosα=1...
cosα=0 => α=Ï€2 => α2=Ï€4 => cosα2=1√2 => secα2 = √2.
Therefore, cosθ.secα2 = 1 x √2 = √2 ....→Option (c)
