If $$\Delta_1 = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$$ and $$\Delta_2 = \begin{vmatrix} x & \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x & 1 \\ \cos 2\theta & 1 & x \end{vmatrix}$$, $$x \neq 0$$; then for all $$\theta \in \left(0, \frac{\pi}{2}\right)$$:

We have two determinants, both of order 3. For any $$3\times 3$$ matrix $$\begin{vmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix}$$ the expansion along the first row is given by

$$\begin{aligned} \text{det}&=a_{11}\bigl(a_{22}a_{33}-a_{23}a_{32}\bigr)\;-\;a_{12}\bigl(a_{21}a_{33}-a_{23}a_{31}\bigr)\;+\;a_{13}\bigl(a_{21}a_{32}-a_{22}a_{31}\bigr). \end{aligned}$$

We shall apply this formula separately to $$\Delta_1$$ and $$\Delta_2$$ and then combine the results.

First, for $$\Delta_1$$ we write

$$\Delta_1=\begin{vmatrix} x & \sin\theta & \cos\theta\\[2pt] -\sin\theta & -x & 1\\[2pt] \cos\theta & 1 & x \end{vmatrix}.$$

Expanding along the first row:

$$\begin{aligned} \Delta_1&=x\;\Bigl[(-x)(x)-1\cdot1\Bigr] -\sin\theta\;\Bigl[(-\sin\theta)(x)-1\cdot\cos\theta\Bigr] +\cos\theta\;\Bigl[(-\sin\theta)(1)-(-x)\cos\theta\Bigr]. \end{aligned}$$

We now evaluate each bracket carefully.

First bracket:

$$(-x)(x)-1\cdot1=-x^2-1.$$

Second bracket:

$$ (-\sin\theta)(x)-1\cdot\cos\theta =-x\sin\theta-\cos\theta. $$

Third bracket:

$$ (-\sin\theta)(1)-(-x)\cos\theta =-\sin\theta+x\cos\theta. $$

Substituting these back:

$$\begin{aligned} \Delta_1 &=x(-x^2-1) -\sin\theta\bigl(-x\sin\theta-\cos\theta\bigr) +\cos\theta\bigl(-\sin\theta+x\cos\theta\bigr). \end{aligned}$$

Distribute the factors in each term:

$$\begin{aligned} \Delta_1 &=-x^3-x +\sin\theta\bigl(x\sin\theta+\cos\theta\bigr) +\cos\theta\bigl(-\sin\theta+x\cos\theta\bigr)\\[4pt] &=-x^3-x +\bigl(x\sin^2\theta+\sin\theta\cos\theta\bigr) +\bigl(-\sin\theta\cos\theta+x\cos^2\theta\bigr). \end{aligned}$$

The mixed terms $$+\sin\theta\cos\theta$$ and $$-\sin\theta\cos\theta$$ cancel out. Adding the remaining terms that contain $$x$$ we observe

$$ x\sin^2\theta+x\cos^2\theta =x\bigl(\sin^2\theta+\cos^2\theta\bigr) =x. $$

So we get

$$ \Delta_1=-x^3-x+x=-x^3. $$

Thus

$$\boxed{\Delta_1=-x^3}. $$

Now we repeat the same procedure for $$\Delta_2$$:

$$\Delta_2=\begin{vmatrix} x & \sin2\theta & \cos2\theta\\[2pt] -\sin2\theta & -x & 1\\[2pt] \cos2\theta & 1 & x \end{vmatrix}.$$

Again expanding along the first row:

$$\begin{aligned} \Delta_2 &=x\;\Bigl[(-x)(x)-1\cdot1\Bigr] -\sin2\theta\;\Bigl[(-\sin2\theta)(x)-1\cdot\cos2\theta\Bigr] +\cos2\theta\;\Bigl[(-\sin2\theta)(1)-(-x)\cos2\theta\Bigr]. \end{aligned}$$

The first bracket is identical to the previous case, still equal to $$-x^2-1$$. The second and third brackets become, exactly as before but with $$2\theta$$ in place of $$\theta$$,

$$ (-\sin2\theta)(x)-1\cdot\cos2\theta=-x\sin2\theta-\cos2\theta, $$ $$ (-\sin2\theta)(1)-(-x)\cos2\theta=-\sin2\theta+x\cos2\theta. $$

Substituting:

$$\begin{aligned} \Delta_2 &=x(-x^2-1) -\sin2\theta\bigl(-x\sin2\theta-\cos2\theta\bigr) +\cos2\theta\bigl(-\sin2\theta+x\cos2\theta\bigr)\\[4pt] &=-x^3-x +\bigl(x\sin^2 2\theta+\sin2\theta\cos2\theta\bigr) +\bigl(-\sin2\theta\cos2\theta+x\cos^2 2\theta\bigr). \end{aligned}$$

Again the terms $$\sin2\theta\cos2\theta$$ cancel. Using $$\sin^2 2\theta+\cos^2 2\theta=1$$ we have

$$ x\sin^2 2\theta+x\cos^2 2\theta=x. $$

Therefore

$$ \Delta_2=-x^3-x+x=-x^3, $$

so that

$$\boxed{\Delta_2=-x^3}. $$

Since both determinants are equal, their sum and their difference are

$$ \Delta_1+\Delta_2=-x^3+(-x^3)=-2x^3, \qquad \Delta_1-\Delta_2=0. $$

Comparing these results with the given options we see that the equality

$$ \Delta_1+\Delta_2=-2x^3 $$

matches Option C (the third option).

Hence, the correct answer is Option C.