If `sin (A – B) =1/2 ? and cos (A + B) = 1/2 " where " A gt B gt 0 and A + B` is an acute angle, then the value B is.
यदि `sin (A - B) =1/2 " और " cos (A + B) =1/2 " जहाँ " A gt B gt 0 " तथा " A+B` न्यून कोण है, तो B का मान क्या होगा?

Options

A. A)`pi/6`
B. B)`pi/12`
C. C)`pi/4`
D. D)`pi/2`

Explanation

To solve the problem, we need to find the value of \( B \) given the equations \( \sin(A - B) = \frac{1}{2} \) and \( \cos(A + B) = \frac{1}{2} \), with the conditions that \( A > B > 0 \) and \( A + B \) is an acute angle.

### Step-by-Step Solution:

1. **Identify the angles for sine and cosine:**
- From \( \sin(A - B) = \frac{1}{2} \), we know that \( A - B \) could be \( 30^\circ \) or \( 150^\circ \) (since \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 150^\circ = \frac{1}{2} \)).
- From \( \cos(A + B) = \frac{1}{2} \), we know that \( A + B \) could be \( 60^\circ \) or \( 300^\circ \) (since \( \cos 60^\circ = \frac{1}{2} \) and \( \cos 300^\circ = \frac{1}{2} \)). However, since \( A + B \) is an acute angle, we only consider \( 60^\circ \).

2. **Set up the equations:**
- From \( A - B = 30^\circ \) (we discard \( 150^\circ \) because \( A > B \) implies \( A - B \) must be positive and less than \( 180^\circ \)).
- From \( A + B = 60^\circ \).

3. **Solve the system of equations:**
- We now have two equations:
\[ A - B = 30^\circ \quad (1) \]
\[ A + B = 60^\circ \quad (2) \]

- Add equations (1) and (2):
\[ (A - B) + (A + B) = 30^\circ + 60^\circ \]
\[ 2A = 90^\circ \]
\[ A = 45^\circ \]

- Now substitute \( A = 45^\circ \) back into equation (2):
\[ 45^\circ + B = 60^\circ \]
\[ B = 60^\circ - 45^\circ = 15^\circ \]

4. **Convert \( B \) to radians:**
- To convert degrees to radians, use the formula:
\[ \text{radians} = \text{degrees} \times \frac{\pi}{180} \]
- Therefore,
\[ B = 15^\circ \times \frac{\pi}{180} = \frac{\pi}{12} \]

### Final Answer:
The value of \( B \) is \( \frac{\pi}{12} \).