This problem requires us to evaluate two $3 \times 3$ determinants, $\Delta_1$ and $\Delta_2$, and then find the relationship between them. The core concept involved is the systematic expansion of a determinant and the application of fundamental trigonometric identities.

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1. Key Concept: Expansion of a $3 \times 3$ Determinant

To evaluate a $3 \times 3$ determinant, we typically expand it along a row or a column. For a general matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, its determinant, denoted as $\det(A)$ or $|A|$, when expanded along the first row, is given by: $\det(A) = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$ Each $2 \times 2$ minor determinant is calculated as $\begin{vmatrix} p & q \\ r & s \end{vmatrix} = ps - qr$. The signs ($+,-,+$) for the terms in the expansion follow a checkerboard pattern starting with $+$ for the $a_{11}$ element.

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2. Step-by-Step Evaluation of $\Delta_1$

Given the determinant: {\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|

Let's expand $\Delta_1$ along its first row: {\Delta _1} = x \cdot \left| {\matrix{ { - x} & 1 \cr 1 & x \cr } } \right| - \sin \theta \cdot \left| {\matrix{ { - \sin \theta } & 1 \cr {\cos \theta } & x \cr } } \right| + \cos \theta \cdot \left| {\matrix{ { - \sin \theta } & { - x} \cr {\cos \theta } & 1 \cr } } \right|

Now, we evaluate each $2 \times 2$ minor determinant:

• The minor for $x$: $(-x)(x) - (1)(1) = -x^2 - 1$
• The minor for $\sin \theta$: $(-\sin \theta)(x) - (1)(\cos \theta) = -x \sin \theta - \cos \theta$
• The minor for $\cos \theta$: $(-\sin \theta)(1) - (-x)(\cos \theta) = -\sin \theta + x \cos \theta$

Substitute these back into the expansion: ${\Delta _1} = x(-x^2 - 1) - \sin \theta (-x \sin \theta - \cos \theta) + \cos \theta (-\sin \theta + x \cos \theta)$

Next, distribute the terms: ${\Delta _1} = -x^3 - x + x \sin^2 \theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2 \theta$

Notice that the terms $+\sin \theta \cos \theta$ and $-\sin \theta \cos \theta$ cancel each other out: ${\Delta _1} = -x^3 - x + x \sin^2 \theta + x \cos^2 \theta$

Factor out $x$ from the trigonometric terms: ${\Delta _1} = -x^3 - x + x (\sin^2 \theta + \cos^2 \theta)$

Apply the fundamental trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$. ${\Delta _1} = -x^3 - x + x(1)$ ${\Delta _1} = -x^3 - x + x$ ${\Delta _1} = -x^3$

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3. Step-by-Step Evaluation of $\Delta_2$

Given the determinant: {\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|

Observe that the structure of $\Delta_2$ is identical to $\Delta_1$, with every instance of $\theta$ replaced by $2\theta$. This means the entire expansion process will follow the same steps. If we substitute $2\theta$ for $\theta$ in the simplified expression for $\Delta_1$, we get: ${\Delta _2} = -x^3 - x + x (\sin^2 2\theta + \cos^2 2\theta)$

Applying the same trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$ (where $\alpha = 2\theta$): ${\Delta _2} = -x^3 - x + x(1)$ ${\Delta _2} = -x^3 - x + x$ ${\Delta _2} = -x^3$

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4. Comparing with Options

We have found that ${\Delta _1} = -x^3$ and ${\Delta _2} = -x^3$. Now let's examine the given options:

• Option (A): ${\Delta _1} - {\Delta _2} = x (\cos 2\theta – \cos 4\theta)$ Substituting our calculated values: $(-x^3) - (-x^3) = 0$ So, option (A) would imply $0 = x (\cos 2\theta – \cos 4\theta)$. Since $x \ne 0$, this would require $\cos 2\theta = \cos 4\theta$ for all $\theta \in (0, \pi/2)$. This is not universally true (e.g., for $\theta = \pi/4$, $\cos(\pi/2) = 0$ and $\cos(\pi) = -1$, which are not equal). Thus, this option is generally incorrect.

• Option (B): ${\Delta _1} + {\Delta _2} = - 2x^3$ Substituting our calculated values: $(-x^3) + (-x^3) = -2x^3$ This matches the option perfectly.

• Option (C): ${\Delta _1} + {\Delta _2} = – 2(x^3 + x –1)$ Substituting our calculated values: $-2x^3 = -2(x^3 + x - 1)$ This simplifies to $x^3 = x^3 + x - 1$, which means $x - 1 = 0$, or $x = 1$. This is not true for all $x \ne 0$. Thus, this option is generally incorrect.

• Option (D): ${\Delta _1} - {\Delta _2} = - 2x^3$ Substituting our calculated values: $(-x^3) - (-x^3) = 0$ So, $0 = -2x^3$. Since $x \ne 0$, this implies $0 = -2(0)$, which means $0=0$. No, this implies $0 = -2x^3$, which is false for $x \ne 0$. Thus, this option is incorrect.

Based on our calculations, Option (B) is the correct relationship.

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5. Tips and Common Mistakes

• Sign Errors: The most common mistake in determinant expansion is incorrect signs. Remember the checkerboard pattern of signs for cofactors: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$.
• Algebraic Simplification: Be careful with distributing negative signs and combining like terms.
• Trigonometric Identities: Always look for opportunities to simplify expressions using fundamental identities like $\sin^2\alpha + \cos^2\alpha = 1$. This often dramatically reduces the complexity of the problem.
• Pattern Recognition: In problems involving similar determinants (like $\Delta_1$ and $\Delta_2$ here), solving one completely often provides a direct path to the solution of the other, saving time and reducing the chance of repeated errors.
• Discrepancy Note: Based on the detailed step-by-step evaluation, $\Delta_1 = -x^3$ and $\Delta_2 = -x^3$. This leads directly to $\Delta_1 + \Delta_2 = -2x^3$ (Option B) and $\Delta_1 - \Delta_2 = 0$. If the provided correct answer is (A), it implies $0 = x(\cos 2\theta - \cos 4\theta)$, which is only true for specific values of $\theta$ (like $\theta = \pi/3$) and not for all $\theta \in (0, \pi/2)$. Therefore, based on standard mathematical evaluation, Option (B) is the logically consistent answer.

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6. Summary and Key Takeaway

This problem demonstrates a straightforward application of determinant expansion for $3 \times 3$ matrices. The key steps involved were:

1. Systematically expanding the determinant along a row or column.
2. Carefully performing algebraic simplifications.
3. Recognizing and applying the fundamental trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$.
4. Leveraging the structural similarity between $\Delta_1$ and $\Delta_2$ to quickly evaluate both.

The final result is that both determinants simplify to $-x^3$, making their sum $-2x^3$. Mastery of determinant evaluation and trigonometric identities is crucial for such problems in JEE Mathematics.

The final answer is $\boxed{\text{B}}$.