If `alpha, beta,gamma,delta` are the solutions of the equation `tan (theta+(pi)/(4))=3tan 3theta`, no two of which have equal tangents.
The value of `(1)/(tan alpha)+(1)/(tan beta)+(1)/(tan gamma)+(1)/(tan delta)` is

Options

A. `-8`
B. `8`
C. `2//3`
D. `1//3`

Explanation

To solve the problem, we need to find the value of \( \frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma} + \frac{1}{\tan \delta} \) given that \( \alpha, \beta, \gamma, \delta \) are the solutions of the equation \( \tan \left( \theta + \frac{\pi}{4} \right) = 3 \tan(3\theta) \).

### Step-by-Step Solution:

1. **Rewrite the Equation**:
We start with the equation:
\[ \tan \left( \theta + \frac{\pi}{4} \right) = 3 \tan(3\theta) \]
Using the tangent addition formula, we have:
\[ \tan \left( \theta + \frac{\pi}{4} \right) = \frac{\tan \theta + \tan \frac{\pi}{4}}{1 - \tan \theta \tan \frac{\pi}{4}} = \frac{\tan \theta + 1}{1 - \tan \theta} \]

2. **Express \( \tan(3\theta) \)**:
The tangent of triple angle can be expressed as:
\[ \tan(3\theta) = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \]
Let \( t = \tan \theta \). Then we can rewrite the equation as:
\[ \frac{t + 1}{1 - t} = 3 \cdot \frac{3t - t^3}{1 - 3t^2} \]

3. **Cross Multiply**:
Cross multiplying gives:
\[ (t + 1)(1 - 3t^2) = 3(3t - t^3)(1 - t) \]
Expanding both sides:
\[ t + 1 - 3t^3 - 3t^2 = 9t - 9t^2 - 3t^3 + 3t^4 \]

4. **Rearranging the Equation**:
Rearranging terms leads to:
\[ 3t^4 - 6t^2 + 8t - 1 = 0 \]

5. **Finding the Roots**:
The polynomial \( 3t^4 - 6t^2 + 8t - 1 = 0 \) has roots \( t_1, t_2, t_3, t_4 \) corresponding to \( \tan \alpha, \tan \beta, \tan \gamma, \tan \delta \).

6. **Using Vieta's Formulas**:
From Vieta's formulas, we can find:
- The sum of the roots \( t_1 + t_2 + t_3 + t_4 = 0 \)
- The sum of the product of the roots taken two at a time \( t_1 t_2 + t_1 t_3 + t_1 t_4 + t_2 t_3 + t_2 t_4 + t_3 t_4 = -2 \)
- The sum of the product of the roots taken three at a time \( t_1 t_2 t_3 + t_1 t_2 t_4 + t_1 t_3 t_4 + t_2 t_3 t_4 = \frac{8}{3} \)
- The product of the roots \( t_1 t_2 t_3 t_4 = \frac{1}{3} \)

7. **Finding the Required Value**:
The expression we need to find is:
\[ \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4} = \frac{t_2 t_3 t_4 + t_1 t_3 t_4 + t_1 t_2 t_4 + t_1 t_2 t_3}{t_1 t_2 t_3 t_4} \]
Substituting the values from Vieta's:
\[ = \frac{\frac{8}{3}}{\frac{1}{3}} = 8 \]

### Final Answer:
The value of \( \frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{1}{\tan \gamma} + \frac{1}{\tan \delta} \) is \( 8 \).