`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 follow these steps:

### Step 1: Apply the formula for the difference of inverse tangents
We use the identity:
\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Here, let \( a = 2 \tan^2 \theta \) and \( b = \frac{1}{3} \tan \theta \).

### Step 2: Substitute \( a \) and \( b \) into the formula
Substituting \( a \) and \( b \) gives us:
\[ \theta = \tan^{-1}\left(\frac{2 \tan^2 \theta - \frac{1}{3} \tan \theta}{1 + 2 \tan^2 \theta \cdot \frac{1}{3} \tan \theta}\right) \]

### Step 3: Simplify the numerator
The numerator simplifies to:
\[ 2 \tan^2 \theta - \frac{1}{3} \tan \theta = \frac{6 \tan^2 \theta - \tan \theta}{3} \]

### Step 4: Simplify the denominator
The denominator simplifies to:
\[ 1 + \frac{2}{3} \tan^3 \theta = \frac{3 + 2 \tan^3 \theta}{3} \]

### Step 5: Combine the fractions
Now, we can combine the fractions:
\[ \theta = \tan^{-1}\left(\frac{\frac{6 \tan^2 \theta - \tan \theta}{3}}{\frac{3 + 2 \tan^3 \theta}{3}}\right) = \tan^{-1}\left(\frac{6 \tan^2 \theta - \tan \theta}{3 + 2 \tan^3 \theta}\right) \]

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

### Step 7: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ \tan \theta (3 + 2 \tan^3 \theta) = 6 \tan^2 \theta - \tan \theta \]

### Step 8: Rearranging the equation
Rearranging leads to:
\[ 2 \tan^4 \theta + 7 \tan \theta - 6 \tan^2 \theta = 0 \]

### Step 9: Factor the equation
Factoring the equation gives:
\[ 2 \tan^3 \theta - 6 \tan^2 \theta + 7 \tan \theta = 0 \]

### Step 10: Solve for \( \tan \theta \)
Factoring out \( \tan \theta \):
\[ \tan \theta (2 \tan^2 \theta - 6 \tan + 7) = 0 \]
This gives us two cases:
1. \( \tan \theta = 0 \)
2. \( 2 \tan^2 \theta - 6 \tan + 7 = 0 \)

### Step 11: Solve the quadratic equation
Using the quadratic formula:
\[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -6, c = 7 \).

### Step 12: Calculate the roots
Calculating the discriminant:
\[ D = (-6)^2 - 4 \cdot 2 \cdot 7 = 36 - 56 = -20 \]
Since the discriminant is negative, there are no real solutions from this quadratic.

### Step 13: Conclusion
Thus, the only solution is \( \tan \theta = 0 \), which corresponds to \( \theta = 0 \).

### Final Answer
The value of \( \tan \theta \) is \( 0 \).

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