If α,β and γ∈0,π2, then the value of sin(α+β+γ)sinα+sinβ+sinγ is
<1
=-1
<0
None of these
Explanation for the correct option:
Find the value of sin(α+β+γ)sinα+sinβ+sinγ:
As we know,
sinα+sinβ+sinγ−sin(α+β+γ)=sinα+sinβ+sinγ−[sinαcos(β+γ)+cosαsin(β+γ)]=sinα+sinβ+sinγ–sinα[cosβcosγ–sinβsinγ]–cosα[sinβcosγ+cosβsinγ]=sinα(1−cosβcosγ)+sinβ(1−cosγcosα)+sinγ(1−cosαcosβ)+sinαsinβsinγ>0
⇒sinα+sinβ+sinγ>sin(α+β+γ) ∵cosα,cosβ,cosγ,sinα,sinβ,sinγ∈[0,1]
∴sin(α+β+γ)sinα+sinβ+sinγ<1
Hence, Option ‘A’ is Correct.