Given, α+β+γ=2π
Now, \Delta = \left|\begin{cases}1 & cos\gamma & cos\beta \\ cos\gamma & 1 & cos\alpha \\ cos\beta & cos\alpha & 1\end{cases}\right|
By expanding along the first column, we get
∆=1(1−cos2α)−cosγ(cosγ−cosα.cosβ)+cosβ(cosα.cosγ−cosβ)
∆=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[cos{2π−(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, ∆=0
Hence, the system of equations has infinitely many solutions.
