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The value of the determinant Δ=∣cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α∣\Delta=\begin{vmatrix}\cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha\\-\sin\beta & \cos\beta & 0\\\sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha\end{vmatrix} is:

Solution

✅ Correct Option: 2

Expanding along the second row and simplifying with sin⁡2+cos⁡2=1\sin^2+\cos^2=1 gives Δ=cos⁡2α+sin⁡2α=1\Delta=\cos^2\alpha+\sin^2\alpha=1. Equivalently, the matrix is a product of two rotation matrices, hence orthogonal with determinant 1.

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