If α,β,γ∈0,π2 then the value of sin(α+β+γ)sinα+sinβ+sinγ is
<1
>1
0
None of these
sin(α+β+γ)=sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγ−sinαsinβsinγ ⇒sin(α+β+γ)−sinα−sinβ−sinγ =sinα(cosβcosγ−1)+sinβ(cosαcosγ−1)+sinγ(cosαcosβ−1)−sinαsinβsinγ ⇒ sin(α+β+γ)−sinα−sinβ−sinγ<0⇒ sin(α+β+γ)<sinα+sinβ+sinγsin(α+β+γ)<1