If `|{:(1+cos^2 theta, sin^2 theta, 2sqrt3tantheta),(cos^2 theta, 1+sin^2 theta, 2sqrt3tan theta),(cos^2 theta, sin^2 theta, 1+2sqrt3 tan theta):}|`=0 then `theta` may be :
i) π/6
ii) 5π/6
iii) 7π/6
iv) 11π/6

Options

A. `pi/6`
B. `(5pi)/6`
C. `(7pi)/6`
D. `(11pi)/6`

Answer

To solve the determinant and find the values of \( \theta \) for which the determinant equals zero, we will follow these steps:

Given the determinant:

\[ D = \begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 2\sqrt{3} \tan \theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 2\sqrt{3} \tan \theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 2\sqrt{3} \tan \theta \end{vmatrix} \]

### Step 1: Apply Row Operations
We will perform the operation \( R_2 \leftarrow R_2 - R_3 \):

\[ D = \begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 2\sqrt{3} \tan \theta \\ \cos^2 \theta - \cos^2 \theta & (1 + \sin^2 \theta - \sin^2 \theta) & (2\sqrt{3} \tan \theta - 2\sqrt{3} \tan \theta) \\ \cos^2 \theta & \sin^2 \theta & 1 + 2\sqrt{3} \tan \theta \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} 1 + \cos^2 \theta & \sin^2 \theta & 2\sqrt{3} \tan \theta \\ 0 & 1 & 0 \\ \cos^2 \theta & \sin^2 \theta & 1 + 2\sqrt{3} \tan \theta \end{vmatrix} \]

### Step 2: Expand the Determinant
Now, we can expand the determinant along the second row:

\[ D = 1 \cdot \begin{vmatrix} 1 + \cos^2 \theta & 2\sqrt{3} \tan \theta \\ \cos^2 \theta & 1 + 2\sqrt{3} \tan \theta \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinant
Calculating the 2x2 determinant:

\[ = (1 + \cos^2 \theta)(1 + 2\sqrt{3} \tan \theta) - (2\sqrt{3} \tan \theta)(\cos^2 \theta) \]

Expanding this gives:

\[ = 1 + 2\sqrt{3} \tan \theta + \cos^2 \theta + 2\sqrt{3} \cos^2 \theta \tan \theta - 2\sqrt{3} \cos^2 \theta \tan \theta \]

This simplifies to:

\[ = 1 + 2\sqrt{3} \tan \theta + \cos^2 \theta \]

### Step 4: Set the Determinant to Zero
Now, we set the determinant to zero:

\[ 1 + 2\sqrt{3} \tan \theta + \cos^2 \theta = 0 \]

### Step 5: Solve for \(\tan \theta\)
Rearranging gives:

\[ 2\sqrt{3} \tan \theta = -1 - \cos^2 \theta \]

Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):

\[ 2\sqrt{3} \tan \theta = -1 - (1 - \sin^2 \theta) = -2 + \sin^2 \theta \]

Thus,

\[ \tan \theta = \frac{-2 + \sin^2 \theta}{2\sqrt{3}} \]

### Step 6: Find Values of \(\theta\)
To find specific values of \( \theta \), we can use the known values of \( \tan \theta \):

From the equation \( \tan \theta = -\frac{1}{2\sqrt{3}} \), we find:

\[ \theta = \tan^{-1}\left(-\frac{1}{2\sqrt{3}}\right) \]

This corresponds to angles in the second and fourth quadrants:

- \( \theta = \frac{5\pi}{6} \) (second quadrant)
- \( \theta = \frac{11\pi}{6} \) (fourth quadrant)

### Final Answer
Thus, the values of \( \theta \) for which the determinant is zero are:

**Options: ii) \( \frac{5\pi}{6} \) and iv) \( \frac{11\pi}{6} \)**.