Question:
Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.
Solution:
$\Delta=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$
$=x\left(-x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)$
$=-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta$
$=-x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$=-x^{3}-x+x$
$=-x^{3}($ Independent of $\theta)$
Hence, $\Delta$ is independent of $\theta$.