The value of the determinant
$\begin{vmatrix}cos\alpha & -sin \alpha & 1\\sin \alpha & cos\alpha & 1\\cos(\alpha + \beta) & - sin (\alpha + \beta ) & 1\end {vmatrix}$ is
Answer & explanation
Correct answer: option 1
The correct answer is option (1) : independent of $\alpha $
We have,
$\begin{vmatrix}cos\alpha & -sin \alpha & 1\\sin \alpha & cos\alpha & 1\\cos(\alpha +\beta) & - sin (\alpha + \beta ) & 1\end {vmatrix}$
$\begin{vmatrix}cos\alpha & -sin \alpha & 1\\sin \alpha & cos\alpha & 1\\0 &0 & 1+sin \beta -cos \beta \end {vmatrix}$ [Applying $R_3→R_3-R_1(cos\beta ) +R_2(sin \beta )]$
$=(1+sin \beta - cos \beta )(cos^2 \alpha + sin^2 \alpha )$
$= 1+ sin \beta -cos\beta , $ which is independent of $\alpha $.