If cos−1α+cos−1β+cos−1γ=3π , then α(β+γ)+β(γ+α)+γ(α+β) equals
0
12
1
6
We have, cos−1α+cos−1β+cos−1γ=3π We know that, 0≤cos−1x≤π⇒cos−1α+cos−1β+cos−1γ=3π is possible if and only if cos−1α=cos−1β=cos−1γ=π⇒α=β=γ=−1∴α(β+γ)+β(γ+α)+γ(α+β)=−1(−1−1)−1(−1−1)−1(−1−1)=6