If `theta_1,theta_2,theta_3` are the three values of `thetain[0,2pi]` for which `tantheta=lambda` then the value of `tan(theta_1)/3tan(theta_2)/3+tan(theta_2)/3tan(theta_3)/3+tan(theta_3)/3tan(theta_1)/3` is equal to

Options

A. `1//3`
B. 1
C. 3
D. none of these

Explanation

To solve the problem, we need to find the value of

\[ \frac{\tan(\theta_1)}{3}\tan(\theta_2) + \frac{\tan(\theta_2)}{3}\tan(\theta_3) + \frac{\tan(\theta_3)}{3}\tan(\theta_1) \]

given that \(\tan(\theta) = \lambda\) for three angles \(\theta_1, \theta_2, \theta_3\) in the interval \([0, 2\pi]\).

### Step 1: Set up the equation
We know that if \(\tan(\theta) = \lambda\), then we can express this in terms of \(\tan(\theta/3)\) using the identity for \(\tan(3\theta)\):

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

### Step 2: Substitute \(\tan(\theta) = \lambda\)
Substituting \(\tan(\theta) = \lambda\) into the equation gives us:

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

### Step 3: Rearranging to form a cubic equation
Rearranging this equation leads to:

\[ 3\tan(\theta) - \tan^3(\theta) = \lambda(1 - 3\tan^2(\theta)) \]

This can be rearranged to form a cubic equation in terms of \(\tan(\theta/3)\):

\[ \tan^3(\theta/3) - 3\lambda\tan^2(\theta/3) - 3\tan(\theta/3) + \lambda = 0 \]

### Step 4: Identify the roots
Let \(x = \tan(\theta/3)\). The roots of this cubic equation are \(x_1 = \tan(\theta_1/3)\), \(x_2 = \tan(\theta_2/3)\), and \(x_3 = \tan(\theta_3/3)\).

### Step 5: Use the sum of products of roots
From Vieta's formulas, the sum of the products of the roots taken two at a time is given by:

\[ x_1 x_2 + x_2 x_3 + x_3 x_1 = -\frac{c}{a} \]

where \(c\) is the coefficient of \(x\) in the cubic equation and \(a\) is the coefficient of \(x^3\). In our case, \(c = -3\) and \(a = 1\), thus:

\[ x_1 x_2 + x_2 x_3 + x_3 x_1 = -(-3) = 3 \]

### Step 6: Calculate the desired expression
Now, we can substitute back into our original expression:

\[ \frac{x_1}{3} x_2 + \frac{x_2}{3} x_3 + \frac{x_3}{3} x_1 = \frac{1}{3}(x_1 x_2 + x_2 x_3 + x_3 x_1) = \frac{1}{3}(3) = 1 \]

### Final Answer
Thus, the value of

\[ \frac{\tan(\theta_1)}{3}\tan(\theta_2) + \frac{\tan(\theta_2)}{3}\tan(\theta_3) + \frac{\tan(\theta_3)}{3}\tan(\theta_1) = 1 \]