If an angle `theta` is divided into two parts A and B such that `A-B=x `and `tanA :tanB=k :1`, then the value of sinx is

Options

A. `(K-1)/(K-1)sinalpha`
B. `(k)/(k+1)sinalpha`
C. `(k-1)/(k+1)sinalpha`
D. none of these

Answer

To solve the problem step by step, we start with the given information and proceed through the necessary trigonometric identities and relationships.

### Step 1: Set up the equations
We are given that:
- \( A - B = x \)
- \( \tan A : \tan B = k : 1 \)

From the ratio of tangents, we can express:
\[ \tan A = k \tan B \]

### Step 2: Express \( \tan A \) and \( \tan B \) in terms of sine and cosine
Using the definition of tangent:
\[ \tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B} \]
Thus, we can write:
\[ \frac{\sin A}{\cos A} = k \cdot \frac{\sin B}{\cos B} \]

### Step 3: Cross-multiply to simplify
Cross-multiplying gives us:
\[ \sin A \cos B = k \sin B \cos A \]

### Step 4: Use the sum and difference identities
We can use the sine addition and subtraction formulas:
- \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
- \( \sin(A - B) = \sin A \cos B - \cos A \sin B \)

From our equation:
\[ \sin A \cos B + \cos A \sin B = k \sin B \cos A \]
Rearranging gives:
\[ \sin A \cos B + \cos A \sin B - k \sin B \cos A = 0 \]

### Step 5: Factor the equation
This can be rewritten as:
\[ \sin A \cos B = (k - 1) \sin B \cos A \]

### Step 6: Substitute \( A + B \) and \( A - B \)
We know that:
- \( A + B = \theta \)
- \( A - B = x \)

Thus, we can express:
\[ \sin(A + B) = \sin \theta, \quad \sin(A - B) = \sin x \]

### Step 7: Relate \( \sin x \) and \( \sin \theta \)
From our earlier rearrangement, we can derive:
\[ \frac{\sin(A - B)}{\sin(A + B)} = \frac{k - 1}{k + 1} \]
Substituting the known values gives:
\[ \frac{\sin x}{\sin \theta} = \frac{k - 1}{k + 1} \]

### Step 8: Solve for \( \sin x \)
Rearranging this gives:
\[ \sin x = \frac{k - 1}{k + 1} \sin \theta \]

### Step 9: Substitute \( \theta \) with \( \alpha \)
If we let \( \theta = \alpha \), we have:
\[ \sin x = \frac{k - 1}{k + 1} \sin \alpha \]

### Final Result
Thus, the value of \( \sin x \) is:
\[ \sin x = \frac{k - 1}{k + 1} \sin \alpha \]