The general value of `theta` satisfying `tantheta tan(120^@-theta) tan(120^@+theta)=1/sqrt3` is

Options

A. `(npi)/(3)-(pi)/(2), n in Z`
B. `(n pi)/(3) - (pi)/(18), n in Z`
C. `(n pi)/(3) + (pi)/(18), n in Z`
D. `(npi)/(3) + (pi)/(12), n in Z`

Answer

To solve the equation \( \tan \theta \tan(120^\circ - \theta) \tan(120^\circ + \theta) = \frac{1}{\sqrt{3}} \), we can follow these steps:

### Step 1: Use the tangent subtraction and addition formulas
We know the formulas for tangent of a difference and a sum:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( A = 120^\circ \) and \( B = \theta \).

### Step 2: Substitute into the equation
Using the formulas, we can express \( \tan(120^\circ - \theta) \) and \( \tan(120^\circ + \theta) \):
\[ \tan(120^\circ - \theta) = \frac{\tan 120^\circ - \tan \theta}{1 + \tan 120^\circ \tan \theta} \]
\[ \tan(120^\circ + \theta) = \frac{\tan 120^\circ + \tan \theta}{1 - \tan 120^\circ \tan \theta} \]

### Step 3: Substitute \( \tan 120^\circ \)
We know that \( \tan 120^\circ = -\sqrt{3} \). Thus, we can substitute this value into our expressions:
\[ \tan(120^\circ - \theta) = \frac{-\sqrt{3} - \tan \theta}{1 - \sqrt{3} \tan \theta} \]
\[ \tan(120^\circ + \theta) = \frac{-\sqrt{3} + \tan \theta}{1 + \sqrt{3} \tan \theta} \]

### Step 4: Substitute back into the original equation
Now substituting these into the original equation:
\[ \tan \theta \cdot \left(\frac{-\sqrt{3} - \tan \theta}{1 - \sqrt{3} \tan \theta}\right) \cdot \left(\frac{-\sqrt{3} + \tan \theta}{1 + \sqrt{3} \tan \theta}\right) = \frac{1}{\sqrt{3}} \]

### Step 5: Simplify the equation
This can be simplified further. The left-hand side becomes:
\[ \tan \theta \cdot \frac{(-\sqrt{3} - \tan \theta)(-\sqrt{3} + \tan \theta)}{(1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta)} \]
The numerator simplifies to:
\[ (-\sqrt{3})^2 - \tan^2 \theta = 3 - \tan^2 \theta \]
And the denominator simplifies to:
\[ 1 - 3 \tan^2 \theta \]
Thus, we have:
\[ \tan \theta \cdot \frac{3 - \tan^2 \theta}{1 - 3 \tan^2 \theta} = \frac{1}{\sqrt{3}} \]

### Step 6: Cross-multiply and rearrange
Cross-multiplying gives:
\[ \tan \theta (3 - \tan^2 \theta) = \frac{1}{\sqrt{3}} (1 - 3 \tan^2 \theta) \]
Expanding both sides:
\[ 3 \tan \theta - \tan^3 \theta = \frac{1}{\sqrt{3}} - \sqrt{3} \tan^2 \theta \]

### Step 7: Rearranging to form a polynomial
Rearranging gives:
\[ \tan^3 \theta - (3 + \sqrt{3}) \tan^2 \theta + 3 \tan \theta - \frac{1}{\sqrt{3}} = 0 \]

### Step 8: Solve the polynomial
This is a cubic equation in \( \tan \theta \). We can find the roots using numerical methods or graphing techniques.

### Step 9: General solution
Once we find \( \theta \), we can express the general solution as:
\[ \theta = n\pi + \text{specific solutions} \]
where \( n \) is any integer.

### Final Answer
After solving the cubic equation, the general solution for \( \theta \) is:
\[ \theta = n\frac{\pi}{3} + \frac{\pi}{18}, \quad n \in \mathbb{Z} \]