If α,β,γ∈(0,π2) then the value of sin(α+β+γ)sinα+sinβ+sinγ is
sin(α+β+γ)=sinαcosβcosγ+sinβcosαcosγ+sinγcosβcosα+sinαsinβsinγ∴sinα+sinβ+sinγ−sin(α+β+γ)=sinα(1−cosβcosγ)+sinβ(1−cosαcosγ)+sinγ(1−cosαcosβ)+sinαsinβsinγ≥0
As α,β,γϵ(0,π2)⇒0
Therefore , sinα+sinβ+sinγ>sinαsinβsinγ⇒sinαsinβsinγsinα+sinβ+sinγ<1
Hence, option 'A' is correct.