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

Options

A. `A=B + npi, n int l`
B. `A=B-npi, n int l`
C. `A=2npi + B, n int I`
D. `A= npi -B, n int l`

Answer

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

1. \( \sin A = \sin B \)
2. \( \cos A = \cos B \)

From these two equations, we can derive the relationship between angles \( A \) and \( B \).

### Step 1: Analyze the equations
Since \( \sin A = \sin B \), we can use the sine identity:
\[ \sin A - \sin B = 0 \]
This can be factored using the identity:
\[ \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) = 0 \]
This implies either:
- \( \cos\left(\frac{A + B}{2}\right) = 0 \) or
- \( \sin\left(\frac{A - B}{2}\right) = 0 \)

### Step 2: Analyze the cosine equation
Similarly, from \( \cos A = \cos B \), we can write:
\[ \cos A - \cos B = 0 \]
Using the cosine identity:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) = 0 \]
This implies either:
- \( \sin\left(\frac{A + B}{2}\right) = 0 \) or
- \( \sin\left(\frac{A - B}{2}\right) = 0 \)

### Step 3: Solve for \( A \) and \( B \)
From both sine and cosine equations, we can conclude that:
1. \( \sin\left(\frac{A - B}{2}\right) = 0 \) implies:
\[ \frac{A - B}{2} = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ A - B = 2n\pi \]
Rearranging gives:
\[ A = 2n\pi + B \]

### Step 4: Identify the correct option
The derived equation \( A = 2n\pi + B \) corresponds to option 3:
- **Option 3**: \( A = 2n\pi + B \) for all values of \( n \in \mathbb{Z} \)

### Conclusion
The correct answer is:
**Option 3: \( A = 2n\pi + B \) for all values of \( n \in \mathbb{Z} \)**.

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