Question

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

• A. independent of $\alpha$
• B. independent of $\beta$
• C. independent of $\alpha$ and $\beta$
• D. none of these

Answer 1

The correct option is A independent of $\alpha$

determinant is $\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1\\ \sin \alpha & \cos \alpha & 1\\ \cos (\alpha +\beta )& -\sin (\alpha +\beta )& 1\end{array}\right|$

By expanding the determinant

$D=(-\sin \alpha \sin (\alpha +\beta )-\cos \alpha \cos (\alpha +\beta \left)\right)-(-\cos \alpha \sin (\alpha +\beta )+\sin \alpha \cos (\alpha +\beta \left)\right)+({\cos }^2\alpha +{\sin }^2\alpha )$

$D=-\cos (\alpha +\beta -\alpha )+\sin (\alpha +\beta -\alpha )+1$

which is equal to $D=\sin \beta -\cos \beta +1$

which is indepedent of $\alpha$