If `A+B=C and tanA=ktanB, and A-B=phi`, then `sinC=?`

Options

A. 0
B. 1
C. `(k+1)/(k-1)`
D. `(k+1)/(k-1)sinphi`

Answer

To solve the problem step by step, we will use the given relationships and trigonometric identities.

### Step 1: Use the given relationships
We have:
1. \( A + B = C \)
2. \( \tan A = k \tan B \)
3. \( A - B = \phi \)

### Step 2: Express \(\tan A\) and \(\tan B\)
Using the definition of tangent:
\[ \tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B} \]
From the second equation, we can write:
\[ \frac{\sin A}{\cos A} = k \cdot \frac{\sin B}{\cos B} \]
This implies:
\[ \sin A \cos B = k \sin B \cos A \]

### Step 3: Apply the sine addition and subtraction formulas
Using the sine addition and subtraction formulas:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
We can rewrite our earlier equation:
\[ \sin A \cos B = k \sin B \cos A \]

### Step 4: Use the component and dividendo method
Using the component and dividendo method:
\[ \frac{a + b}{a - b} = \frac{c + d}{c - d} \]
We can relate this to our sine equations:
\[ \frac{\sin(A + B)}{\sin(A - B)} = \frac{k + 1}{k - 1} \]

### Step 5: Substitute the known values
Substituting \(A + B = C\) and \(A - B = \phi\):
\[ \frac{\sin C}{\sin \phi} = \frac{k + 1}{k - 1} \]

### Step 6: Solve for \(\sin C\)
Rearranging the equation gives:
\[ \sin C = \frac{k + 1}{k - 1} \sin \phi \]

### Final Answer
Thus, the value of \(\sin C\) is:
\[ \sin C = \frac{k + 1}{k - 1} \sin \phi \]

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