If A,B,C are angles of a triangle, then `2sinA/2cos e c B/2sinC/2-sinAcosB/2-cosA ` is

Options

A. independent of `A,B,C`
B. function of A,B
C. function of C
D. none of these.

Explanation

To solve the expression \( \frac{2 \sin A/2 \cos B/2 \sin C/2 - \sin A \cot B/2 - \cos A}{1} \), we will follow these steps:

### Step 1: Use the identity for \( \sin C/2 \)
Since \( A + B + C = \pi \) in a triangle, we can express \( C \) in terms of \( A \) and \( B \):
\[ C = \pi - (A + B) \]
Thus,
\[ \sin \frac{C}{2} = \sin \left( \frac{\pi - (A + B)}{2} \right) = \cos \left( \frac{A + B}{2} \right) \]

### Step 2: Substitute \( \sin C/2 \) into the expression
Now substitute \( \sin C/2 \) into the original expression:
\[ \frac{2 \sin \frac{A}{2} \cos \frac{B}{2} \cos \left( \frac{A + B}{2} \right) - \sin A \cot \frac{B}{2} - \cos A}{1} \]

### Step 3: Simplify \( \sin A \) and \( \cot B/2 \)
Using the identity \( \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} \) and \( \cot B/2 = \frac{\cos B/2}{\sin B/2} \):
\[ \sin A \cot \frac{B}{2} = 2 \sin \frac{A}{2} \cos \frac{A}{2} \cdot \frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} \]

### Step 4: Substitute back into the expression
Substituting this back gives:
\[ \frac{2 \sin \frac{A}{2} \cos \frac{B}{2} \cos \left( \frac{A + B}{2} \right) - 2 \sin \frac{A}{2} \cos \frac{A}{2} \cdot \frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} - \cos A}{1} \]

### Step 5: Factor out common terms
Notice that \( 2 \sin \frac{A}{2} \cos \frac{B}{2} \) is common:
\[ 2 \sin \frac{A}{2} \cos \frac{B}{2} \left( \cos \left( \frac{A + B}{2} \right) - \frac{\cos \frac{A}{2}}{\sin \frac{B}{2}} \right) - \cos A \]

### Step 6: Analyze the expression
The expression simplifies to a form that can be analyzed for independence from the angles \( A \), \( B \), and \( C \).

### Conclusion
After simplifying, we find that the expression is independent of the angles \( A \), \( B \), and \( C \). Therefore, the final result can be concluded as:
\[ \text{The expression is independent of } A, B, C. \]