If `alpha, beta, gamma, delta` satisfy the equation `tan(x+(pi)/(4))=3 tan3x`, then `tanalpha+tanbeta+tangamma+tandelta` is equal to :

Options

A. -1
B. `1//3`
C. 0
D. 2

Answer

To solve the equation \( \tan\left(x + \frac{\pi}{4}\right) = 3 \tan(3x) \) and find \( \tan \alpha + \tan \beta + \tan \gamma + \tan \delta \), we will follow these steps:

### Step 1: Apply the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan\left(x + \frac{\pi}{4}\right) = \frac{\tan x + \tan\frac{\pi}{4}}{1 - \tan x \tan\frac{\pi}{4}} \]
Since \( \tan\frac{\pi}{4} = 1 \), this simplifies to:
\[ \tan\left(x + \frac{\pi}{4}\right) = \frac{\tan x + 1}{1 - \tan x} \]

### Step 2: Write the right-hand side using the tangent triple angle formula
The right-hand side can be expressed using the tangent triple angle formula:
\[ \tan(3x) = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \]
Thus, we have:
\[ 3 \tan(3x) = 3 \cdot \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = \frac{9\tan x - 3\tan^3 x}{1 - 3\tan^2 x} \]

### Step 3: Set the two sides equal
Now we set the two expressions equal:
\[ \frac{\tan x + 1}{1 - \tan x} = \frac{9\tan x - 3\tan^3 x}{1 - 3\tan^2 x} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ (\tan x + 1)(1 - 3\tan^2 x) = (9\tan x - 3\tan^3 x)(1 - \tan x) \]

### Step 5: Expand both sides
Expanding both sides:
Left side:
\[ \tan x + 1 - 3\tan^3 x - 3\tan^2 x \]
Right side:
\[ 9\tan x - 9\tan^2 x - 3\tan^3 x + 3\tan^4 x \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ 3\tan^4 x - 6\tan^2 x + 8\tan x - 1 = 0 \]

### Step 7: Identify the polynomial
This is a polynomial of degree 4, which means it has 4 roots, denoted as \( \tan \alpha, \tan \beta, \tan \gamma, \tan \delta \).

### Step 8: Use Vieta's formulas
According to Vieta's formulas, the sum of the roots of the polynomial \( ax^4 + bx^3 + cx^2 + dx + e = 0 \) is given by \( -\frac{b}{a} \). In our case:
- \( a = 3 \)
- \( b = 0 \) (since there is no \( \tan^3 x \) term)

Thus:
\[ \tan \alpha + \tan \beta + \tan \gamma + \tan \delta = -\frac{0}{3} = 0 \]

### Final Answer
Therefore, the value of \( \tan \alpha + \tan \beta + \tan \gamma + \tan \delta \) is:
\[ \boxed{0} \]