Find the solution of `tan^(-1) 2x +tan^(-1) 3x = pi/4 , x gt 0 `

Answer

To solve the equation \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \) for \( x > 0 \), we can follow these steps:

### Step 1: Use the Addition Formula for Inverse Tangents
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
for \( ab < 1 \). Here, let \( a = 2x \) and \( b = 3x \).

### Step 2: Apply the Formula
Applying the formula, we have:
\[ \tan^{-1}(2x) + \tan^{-1}(3x) = \tan^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}\right) = \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) \]
So, we can rewrite the equation as:
\[ \tan^{-1}\left(\frac{5x}{1 - 6x^2}\right) = \frac{\pi}{4} \]

### Step 3: Take the Tangent of Both Sides
Taking the tangent of both sides gives:
\[ \frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right) = 1 \]

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

### Step 5: Rearrange to Form a Quadratic Equation
Rearranging the equation leads to:
\[ 6x^2 + 5x - 1 = 0 \]

### Step 6: Solve the Quadratic Equation
Now we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6, b = 5, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot (-1) = 25 + 24 = 49 \]

Now substituting into the quadratic formula:
\[ x = \frac{-5 \pm \sqrt{49}}{2 \cdot 6} = \frac{-5 \pm 7}{12} \]

### Step 7: Calculate the Roots
Calculating the two possible values for \( x \):
1. \( x = \frac{-5 + 7}{12} = \frac{2}{12} = \frac{1}{6} \)
2. \( x = \frac{-5 - 7}{12} = \frac{-12}{12} = -1 \)

### Step 8: Consider the Condition \( x > 0 \)
Since we are given that \( x > 0 \), we discard \( x = -1 \). Thus, the only valid solution is:
\[ x = \frac{1}{6} \]

### Final Answer
The solution is:
\[ \boxed{\frac{1}{6}} \]