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 :

Options

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

Answer

To solve the determinant equation given in the question, we need to find the values of \(\theta\) such that:

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

### Step 1: Write the Determinant
The determinant can be expressed as:

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

### Step 2: Apply Row Operations
We can simplify the determinant by performing row operations. Specifically, we can replace the second and third rows with their differences from the first row:

- \(R_2 \rightarrow R_2 - R_1\)
- \(R_3 \rightarrow R_3 - R_1\)

This gives us:

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

This simplifies to:

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

### Step 3: Calculate the Determinant
Now we can calculate the determinant using the first row:

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

Calculating the 2x2 determinants:

1. The first determinant is \(1 \cdot (2\sqrt{3} \tan \theta) - 0 = 2\sqrt{3} \tan \theta\).
2. The second determinant is \(-\cos^2 \theta \cdot (2\sqrt{3} \tan \theta) - 0 = -2\sqrt{3} \cos^2 \theta \tan \theta\).

So we have:

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

### Step 4: Set the Determinant to Zero
Setting \(D = 0\):

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

Factoring out \(2\sqrt{3} \tan \theta\):

\[ 2\sqrt{3} \tan \theta \left(1 + \cos^2 \theta + \sin^2 \theta \cos^2 \theta\right) = 0 \]

### Step 5: Solve for \(\theta\)
This gives us two cases:

1. \(2\sqrt{3} \tan \theta = 0 \Rightarrow \tan \theta = 0 \Rightarrow \theta = n\pi\) (where \(n\) is an integer).
2. The second factor must be zero, which simplifies to \(1 + \cos^2 \theta + \sin^2 \theta \cos^2 \theta = 0\).

However, since \(\cos^2 \theta + \sin^2 \theta = 1\), we can substitute and simplify further.

### Final Values of \(\theta\)
From the first case, we find that:

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

This occurs at:

\[ \theta = \frac{5\pi}{6}, \frac{11\pi}{6} \]

### Conclusion
Thus, the values of \(\theta\) that satisfy the determinant equation are:

\[ \theta = \frac{5\pi}{6}, \frac{11\pi}{6} \]