If α,β,γϵ(0,π2), then sin(α+β+γ)sinα+sinβ+sinγ is-
A
<1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
>1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
<0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B<1 We know, −1<sin(α+β+γ)<1,sin(α+β+γ)=sin(α)cos(β)cos(γ)+cos(α)sin(β)cos(γ)+cos(α)cos(β)sin(γ)−sin(α)sin(β)sin(γ)−3<sinα+sinβ+sinγ<3,∴sin(α+β+γ)sinα+sinβ+sinγ<1