If ` tan ^(-1) 2x + tan ^(-1) 3 xx = (pi)/(4)`, then x = ?

Options

A. 1
B. -1
C. `-(1)/(6)`
D. `(1)/(6)`

Answer

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \), we can use the identity for the sum of inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

### Step-by-Step Solution:

1. **Apply the Identity**:
We apply the identity with \( a = 2x \) and \( b = 3x \):
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]

2. **Set the Equation**:
Since we know that \( \tan^{-1}(1) = \frac{\pi}{4} \), we can set the two expressions equal to each other:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = \tan^{-1}(1) \]

3. **Remove the Inverse Tangent**:
By applying the tangent function to both sides, we get:
\[ \frac{5x}{1 - 6x^2} = 1 \]

4. **Cross-Multiply**:
Cross-multiplying gives:
\[ 5x = 1 - 6x^2 \]

5. **Rearrange the Equation**:
Rearranging this equation leads to:
\[ 6x^2 + 5x - 1 = 0 \]

6. **Solve the Quadratic Equation**:
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6, b = 5, c = -1 \):
\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]
\[ x = \frac{-5 \pm \sqrt{25 + 24}}{12} \]
\[ x = \frac{-5 \pm \sqrt{49}}{12} \]
\[ x = \frac{-5 \pm 7}{12} \]

7. **Calculate the Roots**:
This gives us two potential solutions:
\[ x = \frac{2}{12} = \frac{1}{6} \quad \text{and} \quad x = \frac{-12}{12} = -1 \]

8. **Select the Valid Solution**:
Since \( x \) must be a non-negative value in the context of the problem, we choose:
\[ x = \frac{1}{6} \]

### Final Answer:
Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{6}} \]