If α, β, γ are three real numbers such that α+β+γ=0, then Δ=| 1 cos γ cos β cos γ 1 cos α cos β cos α 1 | text equals

Question

If α, β, &gamma; are three real numbers such that α+β+&gamma;=0, then Δ=| 1 cos &gamma; cos β cos &gamma; 1 cos α cos β cos α 1 | text equals  (A) -1 (B) 0 (C) 1 (D)  cos α cos β cos &gamma;

Tardigrade Admin · Accepted Answer

Let A, B and C be three real numbers such that α=B-C, β=C-A and &gamma;=A-B, clearly α+β+&gamma; =0. We have Δ=|1 cos (A-B) cos (C-A) cos (A-B) 1 cos (B-C) cos (C-A) cos (B-C) 1| | cos2A + sin 2 A& cos A cos B + sin A sin B& cos A cos A + sin A sin C cos A cos B + sin A sin B& cos2B + sin2 B& cos B cos C + sin B sin C cos C cos A + sin C sin A& cos B cos C + sin B sin C& cos2 C + sin 2C| =| cos A sin A 0 cos B sin B 0 cos C sin C 0|| cos A sin A 0 cos B sin B 0 cos C sin C 0|=(0)(0)=0.

Q. If $α, β, γ$ are three real numbers such that $α + β + γ = 0$ , then
$Δ = ∣ ∣ 1 cos γ cos β cos γ 1 cos α cos β cos α 1 ∣ ∣ equals$

-1

0

1

$cos α cos β cos γ$

Q. If α,β,γ are three real numbers such that α+β+γ=0, then Δ=∣∣​1cosγcosβ​cosγ1cosα​cosβcosα1​∣∣​ equals

-1

0

1

cosαcosβcosγ

Q. If $α, β, γ$ are three real numbers such that $α + β + γ = 0$ , then
$Δ = ∣ ∣ 1 cos γ cos β cos γ 1 cos α cos β cos α 1 ∣ ∣ equals$

$cos α cos β cos γ$