If `a^(2)+b^(2)+c^(2)=1` then
`|(a^(2)+(b^(2)+c^(2))costheta, ab(1-cos theta),ac(1-cos theta)),(ba(1-cos theta),b^(2)+(c^(2)+a^(2))cos theta,bc(1-cos theta)),(ca(1-cos theta),cb(1-cos theta),c^(2)+(a^(2)+b^(2))costheta)|` equals

Options

A. `cos^(2)theta`
B. `0`
C. `1`
D. `sin^(2) theta`

Answer

To solve the given determinant problem step by step, we will follow the structure of the determinant and apply the necessary algebraic manipulations.

### Given:
We need to evaluate the determinant:

\[ D = \begin{vmatrix} a^2 + (b^2 + c^2) \cos \theta & ab(1 - \cos \theta) & ac(1 - \cos \theta) \\ ba(1 - \cos \theta) & b^2 + (c^2 + a^2) \cos \theta & bc(1 - \cos \theta) \\ ca(1 - \cos \theta) & cb(1 - \cos \theta) & c^2 + (a^2 + b^2) \cos \theta \end{vmatrix} \]

### Step 1: Simplify the determinant
We know that \( a^2 + b^2 + c^2 = 1 \). Thus, we can rewrite the elements of the determinant:

- The first element in the first row becomes \( a^2 + (1 - a^2) \cos \theta = a^2 + \cos \theta - a^2 \cos \theta = a^2(1 - \cos \theta) + \cos \theta \).
- Similarly, we can simplify the other elements.

### Step 2: Factor out common terms
We can factor out \( a \), \( b \), and \( c \) from the respective rows:

\[ D = abc \begin{vmatrix} \frac{a^2(1 - \cos \theta) + \cos \theta}{a} & \frac{ab(1 - \cos \theta)}{b} & \frac{ac(1 - \cos \theta)}{c} \\ \frac{ba(1 - \cos \theta)}{a} & \frac{b^2(1 - \cos \theta) + \cos \theta}{b} & \frac{bc(1 - \cos \theta)}{c} \\ \frac{ca(1 - \cos \theta)}{a} & \frac{cb(1 - \cos \theta)}{b} & \frac{c^2(1 - \cos \theta) + \cos \theta}{c} \end{vmatrix} \]

### Step 3: Further simplify the determinant
Now, we can simplify the determinant further by substituting \( a^2 + b^2 + c^2 = 1 \):

\[ D = abc \begin{vmatrix} \cos \theta & ab(1 - \cos \theta) & ac(1 - \cos \theta) \\ ba(1 - \cos \theta) & \cos \theta & bc(1 - \cos \theta) \\ ca(1 - \cos \theta) & cb(1 - \cos \theta) & \cos \theta \end{vmatrix} \]

### Step 4: Apply row operations
We can perform row operations to simplify the determinant. For example, we can replace the second and third rows by subtracting the first row from them:

\[ D = abc \begin{vmatrix} \cos \theta & ab(1 - \cos \theta) & ac(1 - \cos \theta) \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 5: Evaluate the determinant
Since the second and third rows are now zero, the determinant evaluates to zero:

\[ D = 0 \]

### Final Result:
Thus, the value of the determinant is:

\[ \boxed{0} \]