`(sin(A-C)+2sinA+sin(A+C))/(sin(B-C)+2sinB+sin(B+C))` is equal to

Options

A. `tan A`
B. `(SinA)/(sinB)`
C. `(cos A)/(cos B)`
D. `(SinC)/(cosB)`

Answer

To solve the expression

\[ \frac{\sin(A - C) + 2\sin A + \sin(A + C)}{\sin(B - C) + 2\sin B + \sin(B + C)}, \]

we will start by simplifying both the numerator and the denominator using the sine addition and subtraction formulas.

### Step 1: Expand the numerator
Using the sine addition and subtraction formulas:

\[ \sin(A - C) = \sin A \cos C - \cos A \sin C, \]
\[ \sin(A + C) = \sin A \cos C + \cos A \sin C. \]

Substituting these into the numerator gives:

\[ \sin(A - C) + \sin(A + C) = (\sin A \cos C - \cos A \sin C) + (\sin A \cos C + \cos A \sin C). \]

### Step 2: Combine like terms in the numerator
Combining the terms, we find:

\[ \sin(A - C) + \sin(A + C) = 2\sin A \cos C. \]

Adding \(2\sin A\) to this result, we have:

\[ 2\sin A \cos C + 2\sin A = 2\sin A(\cos C + 1). \]

### Step 3: Expand the denominator
Now we will do the same for the denominator:

\[ \sin(B - C) = \sin B \cos C - \cos B \sin C, \]
\[ \sin(B + C) = \sin B \cos C + \cos B \sin C. \]

Substituting these into the denominator gives:

\[ \sin(B - C) + \sin(B + C) = (\sin B \cos C - \cos B \sin C) + (\sin B \cos C + \cos B \sin C). \]

### Step 4: Combine like terms in the denominator
Combining the terms, we find:

\[ \sin(B - C) + \sin(B + C) = 2\sin B \cos C. \]

Adding \(2\sin B\) to this result, we have:

\[ 2\sin B \cos C + 2\sin B = 2\sin B(\cos C + 1). \]

### Step 5: Substitute back into the expression
Now substituting the simplified numerator and denominator back into the expression gives:

\[ \frac{2\sin A(\cos C + 1)}{2\sin B(\cos C + 1)}. \]

### Step 6: Simplify the expression
The \(2\) and \((\cos C + 1)\) terms cancel out (assuming \(\cos C + 1 \neq 0\)), leading to:

\[ \frac{\sin A}{\sin B}. \]

### Final Answer
Thus, the expression simplifies to:

\[ \frac{\sin A}{\sin B}. \]