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