If `(tan3A)/(tanA)=k,` then `(sin3A)/(sinA)=`

Options

A. `(3k)/(k-1), k inR`
B. `(2k)/(k-1), k in[1//3,3]`
C. `(2k)/(k-1), k !in[1//3,3]`
D. `(k-1)/(2k), k !in[1//3,3]`

Answer

To solve the problem where \(\frac{\tan 3A}{\tan A} = k\), we need to find \(\frac{\sin 3A}{\sin A}\).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \frac{\tan 3A}{\tan A} = k \]

2. **Recall the definition of tangent:**
\[ \tan A = \frac{\sin A}{\cos A} \quad \text{and} \quad \tan 3A = \frac{\sin 3A}{\cos 3A} \]

3. **Substitute these definitions into the equation:**
\[ \frac{\frac{\sin 3A}{\cos 3A}}{\frac{\sin A}{\cos A}} = k \]

4. **Simplify the equation:**
\[ \frac{\sin 3A \cdot \cos A}{\sin A \cdot \cos 3A} = k \]

5. **Cross-multiply to eliminate the fraction:**
\[ \sin 3A \cdot \cos A = k \cdot \sin A \cdot \cos 3A \]

6. **Rearranging gives:**
\[ \sin 3A \cdot \cos A - k \cdot \sin A \cdot \cos 3A = 0 \]

7. **Using the sine addition and subtraction formulas:**
\[ \sin(3A + A) = \sin 3A \cdot \cos A + \sin A \cdot \cos 3A \]
\[ \sin(3A - A) = \sin 3A \cdot \cos A - \sin A \cdot \cos 3A \]

8. **Thus, we can write:**
\[ \sin 4A = \sin 3A \cdot \cos A + \sin A \cdot \cos 3A \]
\[ \sin 2A = \sin 3A \cdot \cos A - \sin A \cdot \cos 3A \]

9. **Now, we can express \(\frac{\sin 4A}{\sin 2A}\):**
\[ \frac{\sin 4A}{\sin 2A} = \frac{k + 1}{k - 1} \]

10. **Using the sine addition formula:**
\[ \sin 4A = 2 \sin 2A \cos 2A \]

11. **Substituting into the equation gives:**
\[ \frac{2 \sin 2A \cos 2A}{\sin 2A} = \frac{k + 1}{k - 1} \]
\[ 2 \cos 2A = \frac{k + 1}{k - 1} \]

12. **Finally, we can express \(\frac{\sin 3A}{\sin A}\):**
\[ \frac{\sin 3A}{\sin A} = \frac{2k}{k - 1} \]

### Final Result:
Thus, we find that:
\[ \frac{\sin 3A}{\sin A} = \frac{2k}{k - 1} \]