The value of the determinant | 1 cos (α-β) cos α cos (α-β) 1 cos β cos α cos β 1 | is

Question

MARKS Content Team · Accepted Answer

Correct answer: (D) 0. On solving the determinant, we have 1 (1-cos^2 β )-cos (α-β)[cos (α-β) - cos α · cos β]+cos α[cos β · cos (α-β) = 1-cos^2 β-cos^2 α-cos^2(α-β) = +2 cos α · cos β · cos (α-β) = 1-cos^2 β-cos^2 α+cos (α-β) = 1-cos^2 β-cos^2 α+cos (α-β) cos (α+β) [cos (α+β)+cos (α-β)-cos (α-β)] = 1-cos^2 β-cos^2 α+cos^2 α · cos^2 β = 1-cos^2 β-cos^2 α (1-cos^2 β ) = 1-cos^2 β · sin^2 α · sin^2 β = (1-cos^2 β )-sin^2 α · sin^2 β-sin^2 α · sin^2 β = .sin^2 β-sin^2 β 1 )=0

The value of the determinant $\left|\begin{array}{ccc} 1 & \cos (\alpha-\beta) & \cos \alpha \\ \cos (\alpha-\beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{array}\right| \text { is }$

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