Q44 / 90 · JEE Main 2021 (31 Aug Shift 2) · Mathematics

If α+β+γ=2π, then the system of equations

x+cosγy+cosβz=0

cosγx+y+cosαz=0

cosβx+cosαy+z=0

has :

  • Correct answer
    A
    infinitely many solutions
  • B
    a unique solution
  • C
    no solution
  • D
    exactly two solutions

Solution

Given, α+β+γ=2π

Now, Δ=1cosγcosβcosγ1cosαcosβcosα1

By expanding along the first column, we get

∆=11-cos2α-cosγcosγ-cosα.cosβ+cosβcosα.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)-cosγcos2π-(α+β)-2cosα.cosβ

=-cos(2π-γ)cos(α-β)-cosγcos(α+β)-2cosα.cosβ

=-cos(2π-γ).cos(α-β)-cosγcosα.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.

Hence, the correct answer is option: 'A'.

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