Given, α+β+γ=2π
Now, Δ=∣1cosγcosβcosγ1cosαcosβcosα1∣
By expanding along the first column, we get
Δ=1(1−cos2α)−cosγ(cosγ−cosα.cosβ)+cosβ(cosα.cosγ−cosβ)
Δ=1−cos2α−cos2γ+cosγ.cosα.cosβ+cosβ.cosα.cosγ−cos2β
=1−cos2α−cos2β−cos2γ+2cosα.cosβ.cosγ
=sin2α−cos2β−cosγ(cosγ−2cosα.cosβ)
=−cos(α+β).cos(α−β)−cosγ[cos2π−(α+β)−2cosα.cosβ]
=−cos(2π−γ)cos(α−β)−cosγ[cos(α+β)−2cosα.cosβ]
=−cos(2π−γ).cos(α−β)−cosγ[cosα.cosβ−sinα.sinβ−2cosα.cosβ]
=−cosγ.cos(α−β)+cosγ.cos(α−β)
=0
So, Δ=0
Hence, the system of equations has infinitely many solutions.