`theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=`

Options

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

Explanation

To solve the equation \( \theta = \tan^{-1}(2 \tan^2 \theta) - \tan^{-1} \left(\frac{1}{3} \tan \theta\right) \), we will use the identity for the difference of inverse tangents.

### Step 1: Use the identity for the difference of inverse tangents
We know that:
\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a - b}{1 + ab} \right) \]
Let \( a = 2 \tan^2 \theta \) and \( b = \frac{1}{3} \tan \theta \). Then we can rewrite the equation as:
\[ \theta = \tan^{-1} \left( \frac{2 \tan^2 \theta - \frac{1}{3} \tan \theta}{1 + 2 \tan^2 \theta \cdot \frac{1}{3} \tan \theta} \right) \]

### Step 2: Simplify the expression
Now, we simplify the numerator and the denominator:
- **Numerator**:
\[ 2 \tan^2 \theta - \frac{1}{3} \tan \theta = \frac{6 \tan^2 \theta - \tan \theta}{3} = \frac{6 \tan^2 \theta - \tan \theta}{3} \]
- **Denominator**:
\[ 1 + 2 \tan^2 \theta \cdot \frac{1}{3} \tan \theta = 1 + \frac{2}{3} \tan^3 \theta = \frac{3 + 2 \tan^3 \theta}{3} \]

Thus, we can rewrite the equation as:
\[ \theta = \tan^{-1} \left( \frac{6 \tan^2 \theta - \tan \theta}{3 + 2 \tan^3 \theta} \right) \]

### Step 3: Take the tangent of both sides
Taking the tangent of both sides gives:
\[ \tan \theta = \frac{6 \tan^2 \theta - \tan \theta}{3 + 2 \tan^3 \theta} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying yields:
\[ \tan \theta (3 + 2 \tan^3 \theta) = 6 \tan^2 \theta - \tan \theta \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 3 \tan \theta + 2 \tan^4 \theta = 6 \tan^2 \theta - \tan \theta \]
\[ 2 \tan^4 \theta - 6 \tan^2 \theta + 4 \tan \theta = 0 \]

### Step 6: Factor out the common term
Factoring out \( \tan \theta \):
\[ \tan \theta (2 \tan^3 \theta - 6 \tan \theta + 4) = 0 \]

### Step 7: Solve for \( \tan \theta \)
This gives us two cases:
1. \( \tan \theta = 0 \)
2. \( 2 \tan^3 \theta - 6 \tan \theta + 4 = 0 \)

### Step 8: Solve the cubic equation
Let \( x = \tan \theta \). The cubic equation becomes:
\[ 2x^3 - 6x + 4 = 0 \]
Dividing the entire equation by 2:
\[ x^3 - 3x + 2 = 0 \]

### Step 9: Factor the cubic equation
We can factor this equation:
\[ (x + 2)(x^2 - 2x + 1) = 0 \]
This gives us:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
and
\[ (x - 1)^2 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 10: Conclusion
Thus, the possible values for \( \tan \theta \) are:
\[ \tan \theta = -2 \quad \text{or} \quad \tan \theta = 1 \]

Since the question asks for the value of \( \tan \theta \), the answer is:
\[ \boxed{-2} \]