The value of the determinant \( \Delta=\begin{vmatrix}\cos\alpha\,\cos\beta & \cos\alpha\,\sin\beta & -\sin\alpha\\-\sin\beta & \cos\beta & 0\\\sin\alpha\,\cos\beta & \sin\alpha\,\sin\beta & \cos\alpha\end{vmatrix} \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Expanding along the third column, which has a zero:
\( \displaystyle \Delta=-\sin\alpha\begin{vmatrix}-\sin\beta & \cos\beta\\\sin\alpha\cos\beta & \sin\alpha\sin\beta\end{vmatrix}-0+\cos\alpha\begin{vmatrix}\cos\alpha\cos\beta & \cos\alpha\sin\beta\\-\sin\beta & \cos\beta\end{vmatrix} \)
The two 2 × 2 determinants, using \( \sin^2\beta+\cos^2\beta=1 \):
\( \displaystyle -\sin\alpha\sin^2\beta-\sin\alpha\cos^2\beta=-\sin\alpha \)
\( \displaystyle \cos\alpha\cos^2\beta+\cos\alpha\sin^2\beta=\cos\alpha \)
So
\( \displaystyle \Delta=-\sin\alpha(-\sin\alpha)+\cos\alpha(\cos\alpha) \)
\( \displaystyle =\sin^2\alpha+\cos^2\alpha \)
\( \displaystyle =1 \)
NCERT (2026-27 edition):
Determinants
