If `sin (A +B) =1`, where `0 lt B lt 45^@`, what is `cos (A -B)` equal to ?

Options

A. `sin2B`
B. `sin B`
C. `cos 2B`
D. `cos B`

Answer

To solve the problem, we need to find the value of \( \cos(A - B) \) given that \( \sin(A + B) = 1 \) and \( 0 < B < 45^\circ \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
We know that \( \sin(A + B) = 1 \). The sine function reaches its maximum value of 1 at \( 90^\circ \). Therefore, we can set:
\[ A + B = 90^\circ \]

2. **Expressing A in terms of B**:
From the equation \( A + B = 90^\circ \), we can express \( A \) as:
\[ A = 90^\circ - B \]

3. **Finding \( A - B \)**:
Now, we need to find \( A - B \):
\[ A - B = (90^\circ - B) - B = 90^\circ - 2B \]

4. **Calculating \( \cos(A - B) \)**:
We can now substitute \( A - B \) into the cosine function:
\[ \cos(A - B) = \cos(90^\circ - 2B) \]

5. **Using the Co-Function Identity**:
We know from trigonometric identities that:
\[ \cos(90^\circ - x) = \sin(x) \]
Therefore:
\[ \cos(90^\circ - 2B) = \sin(2B) \]

6. **Final Result**:
Thus, we have:
\[ \cos(A - B) = \sin(2B) \]

### Conclusion:
The value of \( \cos(A - B) \) is \( \sin(2B) \).