For α,β,γ≠0, if sin-1α+sin-1β+sin-1γ=π and (α+β+γ)(α-γ+β)=3αβ, then γ equals [2024]
32
3
12
3-122
(1)
Given, sin-1α+sin-1β+sin-1γ=π
and (α+β+γ)(α+β-γ)=3αβ
Let sin-1α=A, sin-1β=B, sin-1γ=C
⇒A+B+C=π and (α+β+γ)(α+β-γ)=3αβ
α2+β2+2αβ-γ2=3αβ, α2+β2-γ2=αβ
α2+β2-γ22αβ=12
⇒cosC=12⇒C=60°⇒sinC=32=γ
So, γ=32