Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.

(A) Let $\Delta = \left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$.
Expanding along the first row:
$\Delta = x(-x^2 - 1) - \sin \theta(-x \sin \theta - \cos \theta) + \cos \theta(-\sin \theta + x \cos \theta)$
$= -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2 \theta$
$= -x^3 - x + x(\sin^2 \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$= -x^3 - x + x(1)$
$= -x^3 - x + x$
$= -x^3$
Since the result $-x^3$ does not contain $\theta$,the determinant is independent of $\theta$.