The correct option is B G.P
A(cosα,sinα),A(cosβ,sinβ),A(cosγ,sinγ),PA2=(cosα+1)2+(sinα−0)2=cos2α+1+2cosα+sin2α=2+2cosα=2(1+cosα)=4cos2(α2)⇒Similarly PB=2cos(β2),PC=2cos(γ2)PA,PB,PC are in G.P⇒2cos(α2),2cos(β2),2cos(γ2) are in G.P⇒cos(α2),cos(β2),cos(γ2) are in G.P.