If `sinA/sinB=sin(A-C)/sin(C-B)` then

Options

A. `b^2,c^2,a^2` are in A.P
B. `a^2,b^2,c^2` are in A.P
C. `c^2,b^2,a^2` are in A.P
D. none

Answer

To solve the equation \(\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}\), we can follow these steps:

### Step 1: Use the Sine Rule
Since \(A\), \(B\), and \(C\) are angles of a triangle, we can use the sine rule, which states:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \]
where \(a\), \(b\), and \(c\) are the sides opposite to angles \(A\), \(B\), and \(C\) respectively.

### Step 2: Express \(\sin A\) and \(\sin B\)
From the sine rule, we can express \(\sin A\) and \(\sin B\) as:
\[ \sin A = ak \quad \text{and} \quad \sin B = bk \]

### Step 3: Substitute into the Equation
Substituting these expressions into the original equation gives:
\[ \frac{ak}{bk} = \frac{\sin(A - C)}{\sin(C - B)} \]
This simplifies to:
\[ \frac{a}{b} = \frac{\sin(A - C)}{\sin(C - B)} \]

### Step 4: Use Angle Sum and Difference Identities
Using the sine difference identity:
\[ \sin(A - C) = \sin A \cos C - \cos A \sin C \]
\[ \sin(C - B) = \sin C \cos B - \cos C \sin B \]
Thus, we can rewrite the right side:
\[ \frac{\sin(A - C)}{\sin(C - B)} = \frac{\sin A \cos C - \cos A \sin C}{\sin C \cos B - \cos C \sin B} \]

### Step 5: Cross Multiply
Cross-multiplying gives:
\[ a(\sin C \cos B - \cos C \sin B) = b(\sin A \cos C - \cos A \sin C) \]

### Step 6: Expand and Rearrange
Expanding both sides:
\[ a \sin C \cos B - a \cos C \sin B = b \sin A \cos C - b \cos A \sin C \]
Rearranging gives:
\[ a \sin C \cos B + b \cos A \sin C = b \sin A \cos C + a \cos C \sin B \]

### Step 7: Analyze the Result
This equation can be analyzed to show that if \(A\), \(B\), and \(C\) are in arithmetic progression, then the equality holds.

### Conclusion
Thus, we conclude that if \(\frac{\sin A}{\sin B} = \frac{\sin(A - C)}{\sin(C - B)}\), then \(A\), \(B\), and \(C\) must be in arithmetic progression.