If `tan3A=(3 tan A+k tan^(3)A)/(1-3 tan^(2)A)`, then k is equal to

Answer

To solve the equation \( \tan 3A = \frac{3 \tan A + k \tan^3 A}{1 - 3 \tan^2 A} \), we need to find the value of \( k \).

### Step-by-Step Solution:

1. **Use the Angle Addition Formula:**
We know that \( \tan 3A \) can be expressed using the formula for \( \tan(A + B) \):
\[ \tan 3A = \tan(A + 2A) \]
Using the formula \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \), we can set \( a = A \) and \( b = 2A \).

2. **Calculate \( \tan 2A \):**
We can express \( \tan 2A \) using the double angle formula:
\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]

3. **Substituting \( \tan 2A \) into \( \tan 3A \):**
Now substituting \( \tan 2A \) into the formula for \( \tan 3A \):
\[ \tan 3A = \frac{\tan A + \tan 2A}{1 - \tan A \tan 2A} \]
Substituting \( \tan 2A \):
\[ \tan 3A = \frac{\tan A + \frac{2 \tan A}{1 - \tan^2 A}}{1 - \tan A \cdot \frac{2 \tan A}{1 - \tan^2 A}} \]

4. **Simplifying the Numerator:**
The numerator becomes:
\[ \tan A + \frac{2 \tan A}{1 - \tan^2 A} = \frac{\tan A(1 - \tan^2 A) + 2 \tan A}{1 - \tan^2 A} = \frac{\tan A(1 - \tan^2 A + 2)}{1 - \tan^2 A} = \frac{\tan A(3 - \tan^2 A)}{1 - \tan^2 A} \]

5. **Simplifying the Denominator:**
The denominator becomes:
\[ 1 - \tan A \cdot \frac{2 \tan A}{1 - \tan^2 A} = 1 - \frac{2 \tan^2 A}{1 - \tan^2 A} = \frac{(1 - \tan^2 A) - 2 \tan^2 A}{1 - \tan^2 A} = \frac{1 - 3 \tan^2 A}{1 - \tan^2 A} \]

6. **Final Expression for \( \tan 3A \):**
Putting it all together:
\[ \tan 3A = \frac{\frac{\tan A(3 - \tan^2 A)}{1 - \tan^2 A}}{\frac{1 - 3 \tan^2 A}{1 - \tan^2 A}} = \frac{\tan A(3 - \tan^2 A)}{1 - 3 \tan^2 A} \]

7. **Comparing with Given Expression:**
Now we compare:
\[ \tan 3A = \frac{3 \tan A + k \tan^3 A}{1 - 3 \tan^2 A} \]
From the expression we derived:
\[ \tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} \]
By comparing coefficients, we see that \( k = -1 \).

### Conclusion:
Thus, the value of \( k \) is:
\[ \boxed{-1} \]