To prove that the determinant is independent of $\theta$, we evaluate the determinant $\Delta = \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix}$. We apply the method of expansion along Row 1 ($R_1$), which is a standard property for evaluating $3 \times 3$ determinants.

Expanding along $R_1$, we get:

$$\Delta = x \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} - \sin \theta \begin{vmatrix} -\sin \theta & 1 \\ \cos \theta & x \end{vmatrix} + \cos \theta \begin{vmatrix} -\sin \theta & -x \\ \cos \theta & 1 \end{vmatrix}$$

Now, we calculate the individual $2 \times 2$ determinants. The first term is $x[(-x)(x) - (1)(1)] = x(-x^2 - 1) = -x^3 - x$. The second term is $-\sin \theta[(-\sin \theta)(x) - (1)(\cos \theta)] = -\sin \theta(-x \sin \theta - \cos \theta) = x \sin^2 \theta + \sin \theta \cos \theta$. The third term is $\cos \theta[(-\sin \theta)(1) - (-x)(\cos \theta)] = \cos \theta(-\sin \theta + x \cos \theta) = -\sin \theta \cos \theta + x \cos^2 \theta$.

Adding these terms together, we simplify the expression:

$$\Delta = (-x^3 - x) + (x \sin^2 \theta + \sin \theta \cos \theta) + (-\sin \theta \cos \theta + x \cos^2 \theta)$$

Grouping like terms, the $\sin \theta \cos \theta$ terms cancel out. We are left with:

$$\Delta = -x^3 - x + x(\sin^2 \theta + \cos^2 \theta)$$

Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, the expression becomes:

$$\Delta = -x^3 - x + x(1) = -x^3 - x + x = -x^3$$

Since the final value of the determinant is $-x^3$, which contains only the variable $x$ and no $\theta$ terms, the determinant is independent of $\theta$. This completes the proof.