If A,B,C are three angles of a triangle , then which one of the following relations is not correct ?

Options

A. sin A = sin (B +C)
B. tan (A-B-C) = tan 2 A
C. `cos ((B+C)/(2)) = sin (A/2)`
D. `cos (B + C)= cos A`

Answer

To determine which relation among the given options is not correct for the angles A, B, and C of a triangle, we can analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding the Triangle Angle Sum Property**:
- In any triangle, the sum of the angles A, B, and C is always 180 degrees.
- Therefore, we have:
\[ A + B + C = 180^\circ \]

2. **Analyzing Option 1**:
- The relation states:
\[ \sin A = \sin(180^\circ - (B + C)) \]
- Since \(B + C = 180^\circ - A\), we can rewrite it as:
\[ \sin A = \sin A \]
- This is a true statement, so **Option 1 is correct**.

3. **Analyzing Option 2**:
- The relation states:
\[ \tan(A - (B + C)) = \tan(2A) \]
- We know that \(B + C = 180^\circ - A\), so:
\[ A - (B + C) = A - (180^\circ - A) = 2A - 180^\circ \]
- Therefore, we need to check:
\[ \tan(2A - 180^\circ) = \tan(2A) \]
- The tangent function has a periodicity of 180 degrees, hence:
\[ \tan(2A - 180^\circ) = \tan(2A) \]
- This is also true, so **Option 2 is correct**.

4. **Analyzing Option 3**:
- The relation states:
\[ \cos\left(\frac{B + C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \]
- Using the fact that \(B + C = 180^\circ - A\):
\[ \frac{B + C}{2} = \frac{180^\circ - A}{2} = 90^\circ - \frac{A}{2} \]
- Thus:
\[ \cos\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A}{2}\right) \]
- This relation holds true, so **Option 3 is correct**.

5. **Analyzing Option 4**:
- The relation states:
\[ \cos(B + C) = \cos(180^\circ - A) \]
- We know that:
\[ \cos(180^\circ - A) = -\cos A \]
- However, since \(B + C = 180^\circ - A\), we have:
\[ \cos(B + C) = \cos(180^\circ - A) = -\cos A \]
- This implies:
\[ \cos(B + C) \neq \cos A \]
- Therefore, **Option 4 is incorrect**.

### Conclusion:
The relation that is not correct is **Option 4**.