The value of `theta` lying between `theta =0 and pi//2` and satisfying the equation `|(1+sin^2theta,cos^2theta,4sin4theta),(sin^2theta,1+cos^2theta,4sin4theta),(sin^2theta,cos^2theta,1+4sin4theta)|=0` are

Options

A. `7pi//24`
B. `5pi//24`
C. `11pi//24`
D. `pi//24`

Answer

To solve the problem, we need to find the values of \( \theta \) in the interval \( (0, \frac{\pi}{2}) \) that satisfy the determinant equation:

\[ \left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| = 0 \]

### Step 1: Simplify the Determinant

We can perform column operations to simplify the determinant. Let's add the first column to the second column:

\[ C_2 \rightarrow C_2 + C_1 \]

This gives us:

\[ \left| \begin{array}{ccc} 1 + \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| \]

### Step 2: Further Simplification

Notice that \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, we can rewrite the determinant as:

\[ \left| \begin{array}{ccc} 1 + \sin^2 \theta & 1 & 4 \sin 4\theta \\ \sin^2 \theta & 1 & 4 \sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| \]

### Step 3: Row Operations

Now, let's perform row operations. Subtract the first row from the second row:

\[ R_2 \rightarrow R_2 - R_1 \]

This gives us:

\[ \left| \begin{array}{ccc} 1 + \sin^2 \theta & 1 & 4 \sin 4\theta \\ \sin^2 \theta - (1 + \sin^2 \theta) & 0 & 0 \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| \]

This simplifies to:

\[ \left| \begin{array}{ccc} 1 + \sin^2 \theta & 1 & 4 \sin 4\theta \\ -\cos^2 \theta & 0 & 0 \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| \]

### Step 4: Expand the Determinant

Now, we can expand the determinant. The second column has a zero, which simplifies our calculations:

\[ = 0 \cdot \text{(some determinant)} - 4 \sin 4\theta \cdot \left| \begin{array}{cc} -\cos^2 \theta & 0 \\ \sin^2 \theta & \cos^2 \theta \end{array} \right| \]

Calculating the 2x2 determinant:

\[ = -\cos^2 \theta \cdot 0 - 0 \cdot \sin^2 \theta = 0 \]

### Step 5: Solve for \( \theta \)

The determinant simplifies to:

\[ 4 \sin 4\theta \cdot \cos^2 \theta = 0 \]

This gives us two cases:

1. \( \sin 4\theta = 0 \)
2. \( \cos^2 \theta = 0 \)

**Case 1: \( \sin 4\theta = 0 \)**

This implies:

\[ 4\theta = n\pi \quad \text{for } n \in \mathbb{Z} \]

Thus,

\[ \theta = \frac{n\pi}{4} \]

In the interval \( (0, \frac{\pi}{2}) \), the valid values of \( n \) are 1 and 2, giving:

\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{\pi}{2} \]

**Case 2: \( \cos^2 \theta = 0 \)**

This implies:

\[ \cos \theta = 0 \implies \theta = \frac{\pi}{2} \]

### Final Values

Thus, the values of \( \theta \) in the interval \( (0, \frac{\pi}{2}) \) that satisfy the determinant equation are:

\[ \theta = \frac{\pi}{4} \]