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

Answer

To solve the problem, we need to find the value of \( k \) in the equation:

\[ \tan 3A = \frac{3 \tan A + k \tan^3 A}{1 - 3 \tan^2 A} \]

### Step 1: Use the angle addition formula for tangent

We can express \( \tan 3A \) as \( \tan(2A + A) \). Using the tangent addition formula:

\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]

we can write:

\[ \tan 3A = \tan(2A + A) = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A} \]

### Step 2: Find \( \tan 2A \)

Using the double angle formula for tangent:

\[ \tan 2A = \frac{2 \tan A}{1 - \tan^2 A} \]

### Step 3: Substitute \( \tan 2A \) into the equation

Now, substitute \( \tan 2A \) into the expression for \( \tan 3A \):

\[ \tan 3A = \frac{\frac{2 \tan A}{1 - \tan^2 A} + \tan A}{1 - \frac{2 \tan A}{1 - \tan^2 A} \tan A} \]

### Step 4: Simplify the numerator

The numerator becomes:

\[ \frac{2 \tan A + \tan A(1 - \tan^2 A)}{1 - \tan^2 A} = \frac{2 \tan A + \tan A - \tan^3 A}{1 - \tan^2 A} = \frac{3 \tan A - \tan^3 A}{1 - \tan^2 A} \]

### Step 5: Simplify the denominator

The denominator simplifies to:

\[ 1 - \frac{2 \tan^2 A}{1 - \tan^2 A} = \frac{(1 - \tan^2 A) - 2 \tan^2 A}{1 - \tan^2 A} = \frac{1 - 3 \tan^2 A}{1 - \tan^2 A} \]

### Step 6: Combine the results

Now we can combine the results:

\[ \tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} \]

### Step 7: Compare with the original equation

From the original equation, we have:

\[ \tan 3A = \frac{3 \tan A + k \tan^3 A}{1 - 3 \tan^2 A} \]

By comparing the numerators, we get:

\[ 3 \tan A - \tan^3 A = 3 \tan A + k \tan^3 A \]

### Step 8: Solve for \( k \)

Now, equate the coefficients of \( \tan^3 A \):

\[ -1 = k \]

Thus, we find that:

\[ k = -1 \]

### Final Answer

The value of \( k \) is:

\[ \boxed{-1} \]