1. Write a general determinant

Let

 a=\sin\phi,\qquad b=\cos\phi

and consider

$$D(\phi)=\begin{vmatrix} x & a & b\\ -a & -x & 1\\ b & 1 & x \end{vmatrix}.$$

Then

$$\Delta_1=D(\theta),\qquad \Delta_2=D(2\theta).$$

We compute $D(\phi)$ once and then substitute.

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2. Expand the determinant along the first row

$$D(\phi)=x\begin{vmatrix}-x & 1\\ 1 & x\end{vmatrix} -a\begin{vmatrix}-a & 1\\ b & x\end{vmatrix} +b\begin{vmatrix}-a & -x\\ b & 1\end{vmatrix}.$$

Now evaluate each minor:

• First minor:

$$\begin{vmatrix}-x & 1\\ 1 & x\end{vmatrix}=(-x)(x)-1=-x^2-1.$$

So its contribution is

$$x(-x^2-1)=-x^3-x.$$

• Second minor:

$$\begin{vmatrix}-a & 1\\ b & x\end{vmatrix}=(-a)(x)-b=-ax-b.$$

So its contribution is

$$-a(-ax-b)=a(ax+b)=a^2x+ab.$$

• Third minor:

$$\begin{vmatrix}-a & -x\\ b & 1\end{vmatrix}=(-a)(1)-(-x)b=-a+bx.$$

So its contribution is

$$b(-a+bx)=-ab+b^2x.$$

Hence,

$$D(\phi)=(-x^3-x)+(a^2x+ab)+(-ab+b^2x).$$

The $ab$ terms cancel:

$$D(\phi)=-x^3-x+x(a^2+b^2).$$

Using

$$a^2+b^2=\sin^2\phi+\cos^2\phi=1,$$

we get

$$D(\phi)=-x^3-x+x=-x^3.$$

Therefore,

$$\Delta_1=-x^3,\qquad \Delta_2=-x^3.$$

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3. Now compute the required expressions

Since both are equal to $-x^3$,

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

and

$$\Delta_1-\Delta_2=0.$$

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4. Check each option

• A:

$$\Delta_1-\Delta_2=x(\cos2\theta-\cos4\theta)$$

But actually $\Delta_1-\Delta_2=0$ for all $\theta$, so this is false in general.

• B:

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

This is true.

• C:

$$\Delta_1+\Delta_2=-2(x^3+x-1)$$

False.

• D:

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

False, because $\Delta_1-\Delta_2=0$.

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5. Final answer

The correct option is:

$$\boxed{\text{B}}$$

This matches the stored correct answer.