Problem 15 - Olympiad

Using the identity cos(α+β) = cos α cos β - sin α sin β, and cos(α+γ) = cos α cos γ - sin α sin γ. Write the determinant as: | cos α cos α cos β - sin α sin β cos α cos γ - sin α sin γ |. Factor cos α from the first column: cos α · | 1 cos β - (tan α) sin β cos γ - (tan α) sin γ |. Subtract (cos β) times column 1 from column 2, and (cos γ) times column 1 from column 3: Column 2 becomes: cos β - tan α sin β - cos β·1 = -tan α sin β. Column 3 becomes: cos γ - tan α sin γ - cos γ·1 = -tan α sin γ. So the determinant becomes: cos α · | 1 -tan α sin β -tan α sin γ |. Factor -tan α from columns 2 and 3: cos α · (-tan α)^2 · | 1 sin β sin γ |. Now compute | 1 sin β sin γ |. Using the identity: | 1 sin β sin γ | = 0 because the third row? Wait, for a 3×3 determinant with first row [1, sin β, sin γ], second row and third row need to be specified. The original matrix is 3×3 with three rows, all having cos α as the first entry. So the matrix is: Row1: [cos α, cos(α+β), cos(α+γ)] Row2: [cos β, cos(β+β?), no — the determinant shown only has one row displayed. The question seems incomplete. Let me assume the full determinant is a 3×3 matrix with rows [cos α, cos(α+β), cos(α+γ)], [cos β, cos(β+β), cos(β+γ)], [cos γ, cos(γ+β), cos(γ+γ)]? This is unclear. Given the options, the standard result for | cos α cos(α+β) cos(α+γ) | | cos β cos(β+β) cos(β+γ) | | cos γ cos(γ+β) cos(γ+γ) | is 0. But I cannot determine the exact matrix from the question as written. Based on typical exam questions, the answer is 0.