If sinA = sinB and cosA=cosB, then which one of the following is correct ?

Options

A. `B=npi+A`
B. `A=2npi-B`
C. `A=2npi+B`
D. `B=npi-A` (n is an integer)

Explanation

To solve the problem, we start with the given conditions:

1. **Given**:
- \( \sin A = \sin B \)
- \( \cos A = \cos B \)

2. **Understanding the implications**:
- If \( \sin A = \sin B \), this implies that \( A \) and \( B \) could either be equal or differ by an integer multiple of \( \pi \) (i.e., \( A = B + 2n\pi \) or \( A = \pi - B + 2n\pi \)).
- If \( \cos A = \cos B \), this implies that \( A \) and \( B \) could either be equal or differ by an integer multiple of \( 2\pi \) (i.e., \( A = B + 2n\pi \) or \( A = -B + 2n\pi \)).

3. **Combining the two conditions**:
- Since both sine and cosine are equal, the only way both conditions can hold true simultaneously is if \( A \) and \( B \) are equal or differ by \( 2n\pi \) (the periodicity of cosine) or differ by \( n\pi \) (the periodicity of sine).
- Therefore, we can conclude that:
\[ A = B + 2n\pi \quad \text{or} \quad A = -B + 2n\pi \]

4. **Using the tangent function**:
- We can also express this in terms of tangent:
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\sin B}{\cos B} = \tan B \]
- This gives us:
\[ \tan A = \tan B \]
- The general solution for \( \tan A = \tan B \) is:
\[ B = A + n\pi \]

5. **Final conclusion**:
- Since both sine and cosine are equal, the relationship can be expressed as:
\[ B = n\pi + A \]
- This indicates that \( B \) is equal to \( A \) plus an integer multiple of \( \pi \).

Thus, the correct answer is:
\[ B = n\pi + A \]