If `alpha, beta` are the roots of the equation `6x^(2) - 5x + 1 = 0 `, then the value of `(1)/(pi) [tan ^(-1) (alpha) + tan^(-1) (beta)]` is ________

Answer

To solve the problem, we need to find the value of \(\frac{1}{\pi} \left( \tan^{-1}(\alpha) + \tan^{-1}(\beta) \right)\) where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(6x^2 - 5x + 1 = 0\).

### Step 1: Find the roots of the quadratic equation
The roots of the quadratic equation \(ax^2 + bx + c = 0\) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \(6x^2 - 5x + 1 = 0\), we have:
- \(a = 6\)
- \(b = -5\)
- \(c = 1\)

Calculating the discriminant:
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 6 \cdot 1 = 25 - 24 = 1 \]
Now substituting into the quadratic formula:
\[ x = \frac{5 \pm \sqrt{1}}{2 \cdot 6} = \frac{5 \pm 1}{12} \]
This gives us the roots:
\[ \alpha = \frac{6}{12} = \frac{1}{2}, \quad \beta = \frac{4}{12} = \frac{1}{3} \]

### Step 2: Use the formula for \(\tan^{-1}(\alpha) + \tan^{-1}(\beta)\)
We can use the identity:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \quad \text{if } xy < 1 \]
Here, \(x = \alpha = \frac{1}{2}\) and \(y = \beta = \frac{1}{3}\).

Calculating \(xy\):
\[ xy = \alpha \beta = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} < 1 \]
Thus, we can apply the formula:
\[ \tan^{-1}(\alpha) + \tan^{-1}(\beta) = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}}\right) \]

### Step 3: Simplify the expression
Calculating the numerator:
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]
Calculating the denominator:
\[ 1 - \frac{1}{6} = \frac{5}{6} \]
Thus, we have:
\[ \tan^{-1}(\alpha) + \tan^{-1}(\beta) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) \]

### Step 4: Find the value of \(\tan^{-1}(1)\)
We know that:
\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 5: Substitute back to find \(X\)
Now substituting back into our expression:
\[ \frac{1}{\pi} \left( \tan^{-1}(\alpha) + \tan^{-1}(\beta) \right) = \frac{1}{\pi} \cdot \frac{\pi}{4} = \frac{1}{4} \]

### Final Answer
Thus, the value of \(\frac{1}{\pi} \left( \tan^{-1}(\alpha) + \tan^{-1}(\beta) \right)\) is:
\[ \boxed{\frac{1}{4}} \]