`|{:(cos ( theta +A), sin ( theta +A),1),(cos ( theta +B),sin ( theta +B),1),(cos ( theta +C),sin ( theta +C ) ,1):}|` is independent of

Options

A. A
B. B
C. C
D. `theta`

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} \cos(\theta + A) & \sin(\theta + A) & 1 \\ \cos(\theta + B) & \sin(\theta + B) & 1 \\ \cos(\theta + C) & \sin(\theta + C) & 1 \end{vmatrix} \]

### Step 1: Expand the Determinant
We will expand the determinant using the formula for a 3x3 determinant:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = \cos(\theta + A) \)
- \( b = \sin(\theta + A) \)
- \( c = 1 \)
- \( d = \cos(\theta + B) \)
- \( e = \sin(\theta + B) \)
- \( f = 1 \)
- \( g = \cos(\theta + C) \)
- \( h = \sin(\theta + C) \)
- \( i = 1 \)

Thus, we can express the determinant as:

\[ D = \cos(\theta + A)(\sin(\theta + B) \cdot 1 - \sin(\theta + C) \cdot 1) - \sin(\theta + A)(\cos(\theta + B) \cdot 1 - \cos(\theta + C) \cdot 1) + 1(\cos(\theta + B) \sin(\theta + C) - \sin(\theta + B) \cos(\theta + C)) \]

### Step 2: Simplify the Terms
Now, we simplify each term:

1. The first term becomes:
\[ \cos(\theta + A)(\sin(\theta + B) - \sin(\theta + C) \]

2. The second term simplifies to:
\[ -\sin(\theta + A)(\cos(\theta + B) - \cos(\theta + C) \]

3. The third term simplifies to:
\[ \cos(\theta + B) \sin(\theta + C) - \sin(\theta + B) \cos(\theta + C) = \sin(C - B) \]

### Step 3: Combine the Terms
Combining all these terms, we have:

\[ D = \cos(\theta + A)(\sin(\theta + B) - \sin(\theta + C)) - \sin(\theta + A)(\cos(\theta + B) - \cos(\theta + C)) + \sin(C - B) \]

### Step 4: Analyze Independence
To determine the independence of the determinant with respect to the variable \( \theta \), we observe that the terms involving \( \theta \) can be expressed in terms of sine and cosine functions. However, the final expression contains only the differences \( A, B, C \) and does not depend on \( \theta \).

Thus, we conclude that the determinant \( D \) is independent of \( \theta \).

### Final Conclusion
The determinant \( D \) is independent of \( \theta \).

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