sin(β+γ-α)+sin(γ+α-β)+sin(α+β-γ)-sin(α+β+γ)=
2sinαsinβsinγ
4sinαsinβsinγ
sinαsinβsinγ
None of these
Explanation for the correct option:
Simplifying the equations using trigonometric identities:
⇒sin(β+γ-α)+sin(γ+α-β)+sin(α+β-γ)-sin(α+β+γ)Usingformulae;sinx-y=sinxcosy-sinycosxsinx+y=sinxcosy+sinycosx⇒sin(β+γ)cosα–sinαcos(β+γ)+sin(α+γ)cosβ–sinβcos(α+γ)+sin(α+β)cosγ–sinγcos(α+β)–sin(α+β)cosγ+sinγcos(α+β)⇒sin(β+γ)cosα–sinαcos(β+γ)+sin(α+γ)cosβ–sinβcos(α+γ)+sin(α+β)cosγ–sinγcos(α+β)–sin(α+β)cosγ–sinγcos(α+β)⇒[sinβcosγ+cosβsinγ]cosα–sinα[cosβcosγ–sinβsinγ]+cosβ[sinαcosγ+cosαsinγ]–sinβ[cosαcosγ–sinαsinγ]–2sinγ[cosαcosβ–sinαsinβ]⇒cosαsinβcosγ+cosαcosβsinγ–sinαcosβcosγ+sinαsinβsinγ+sinαcosβcosγ+cosαcosβsinγ–cosαsinβcosγ+sinαsinβsinγ–2cosαcosβsinγ+2sinαsinβsinγ⇒–2cosαcosβsinγ
Therefore, the correct answer is option (D).
Using properties of determinants, prove that: