The parameter, on which the value of the determinant |1 a a²; cos(p-d)x cos px cos(p+d)x; sin(p-d)x

Q: Why did we choose $C_1 + C_3$ instead of just expanding the determinant directly?

Expanding a $3 \times 3$ determinant directly is a classic trap! It leads to a massive algebraic mess that is highly prone to calculation errors. By using $C_1 + C_3$, we exploit the symmetry of the AP angles, which simplifies the expression before we even touch the expansion. In JEE, always look for symmetry before you start calculating.

Q: How do I know which column to operate on?

Look for the 'middle' term. Here, $px$ is the average of $(p-d)x$ and $(p+d)x$. This tells us that the middle column ($C_2$) is the pivot. We want to manipulate $C_1$ and $C_3$ to match or eliminate terms relative to $C_2$. The middle column is your anchor.

Q: Is it possible to solve this by substituting values for $p, d, x$?

Yes! This is a classic JEE 'smart' trick. If the problem states the determinant is independent of $p$, you can set $p=0$ or any convenient value to simplify the calculation. However, understanding the algebraic derivation is crucial for your conceptual growth and will help you solve more complex variations of this problem.

Question

The parameter, on which the value of the determinant $\begin{vmatrix} 1 & a & a^2 \ \cos(p-d)x & \cos px & \cos(p+d)x \ \sin(p-d)x & \sin px & \sin(p+d)x \end{vmatrix}$ does not depend upon is

Options:

(A) (a) $a$

(B) (b) $p$

(C) (c) $d$

(D) (d) $x$

Sigma Percentile · Accepted Answer

### The Art of Seeing Symmetry: Unlocking the Determinant

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of trigonometry and matrix algebra. You see a $3 	imes 3$ determinant, and your instinct might be to expand it immediately.

Stop. Take a breath. In the world of JEE Advanced, brute force is rarely the intended path. There is almost always a hidden elegance, a structural beauty waiting to be uncovered. Let us embark on this journey together.

### Phase 1: The Observation

Let us define our determinant as $\Delta$:

$$\Delta = \begin{vmatrix} 1 & a & a^2 \ \cos(p-d)x & \cos px & \cos(p+d)x \ \sin(p-d)x & \sin px & \sin(p+d)x \end{vmatrix}$$

Look at the second and third rows. Do you see it? The angles are $(p-d)x$, $px$, and $(p+d)x$. These are in an Arithmetic Progression (AP).

Whenever you see an AP in a determinant, it is a signal from the examiner. It is a whisper saying, 'Use column operations.' The symmetry here is not accidental; it is the key to the entire problem.

### Phase 2: The Operation

We want to simplify these trigonometric terms. We have $C_1$ containing $(p-d)x$ and $C_3$ containing $(p+d)x$. If we add them, we can invoke the sum-to-product identities.

Let us perform the operation $C_1 ightarrow C_1 + C_3$:

$$\Delta = \begin{vmatrix} 1+a^2 & a & a^2 \ \cos(p-d)x + \cos(p+d)x & \cos px & \cos(p+d)x \ \sin(p-d)x + \sin(p+d)x & \sin px & \sin(p+d)x \end{vmatrix}$$

Now, recall your trigonometric toolkit. We know that $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$ and $\sin(A-B) + \sin(A+B) = 2\sin A \sin B$. By setting $A = px$ and $B = dx$, our determinant transforms into something much cleaner:

$$\Delta = \begin{vmatrix} 1+a^2 & a & a^2 \ 2\cos px \cos dx & \cos px & \cos(p+d)x \ 2\sin px \cos dx & \sin px & \sin(p+d)x \end{vmatrix}$$

### Phase 3: The Kill

Look at the first column and the second column. Do you see the commonality? The first column has $2\cos px \cos dx$ and $2\sin px \cos dx$, while the second column has $\cos px$ and $\sin px$.

They are almost identical, differing only by a factor of $2\cos dx$. This is the moment of truth. We can create zeros in the first column to make the expansion trivial.

Let us apply $C_1 ightarrow C_1 - (2\cos dx)C_2$:

$$\Delta = \begin{vmatrix} 1+a^2-2a\cos dx & a & a^2 \ 0 & \cos px & \cos(p+d)x \ 0 & \sin px & \sin(p+d)x \end{vmatrix}$$

With two zeros in the first column, the expansion is no longer a chore; it is a victory. We expand along the first column:

$$\Delta = (1+a^2-2a\cos dx) \begin{vmatrix} \cos px & \cos(p+d)x \ \sin px & \sin(p+d)x \end{vmatrix}$$

### Phase 4: The Final Elegance

Now, we evaluate the remaining $2 	imes 2$ determinant. This is standard cross-multiplication:

$$\Delta = (1+a^2-2a\cos dx) [\cos px \sin(p+d)x - \sin px \cos(p+d)x]$$

This expression inside the bracket is the sine subtraction formula, $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Here, $A = (p+d)x$ and $B = px$. Thus, the bracket simplifies to $\sin((p+d)x - px) = \sin(dx)$.

Our final result is:

$$\Delta = (1+a^2-2a\cos dx) \sin(dx)$$

Look at the result. Where is $p$? It has vanished! It has been cancelled out by the beauty of the trigonometric identities. The determinant is entirely independent of $p$.

This is the power of mathematical manipulation—taking a complex, intimidating structure and revealing the simple, elegant truth hidden underneath. You have successfully navigated the trap. **Well done.**

Animated Solution for Mathematics - Matrices and Determinants: The parameter, on which the value of the determinant $1 cos (p - d) x sin (p - d) x a cos p x sin p x a^{2} cos (p + d) x sin (p + d) x$ does not depend upon is

Select Answer:

Visualized Solution

Defining the Determinant $Δ$

Analyzing Column Relationships

Applying Column Operation $C_{1} \to C_{1} + C_{3}$

Trigonometric Sum-to-Product Identities

Simplifying Column 1

Creating Zeros in Column 1

Expanding Along Column $C_{1}$

Evaluating the $2 \times 2$ Determinant

Final Simplification and Conclusion

The Sigma Insight: Properties of Determinants

The Art of Seeing Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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