If in a triangle ABC ,
`2(cosA)/a+(cosB)/b+2(cosC)/c=a/(bc)+b/(ca)`,
then the value of the angle A, is

Options

A. `pi/3`
B. `pi/4`
C. `pi/2`
D. `pi/6`

Answer

To solve the problem, we start with the given equation in triangle ABC:

\[ 2 \frac{\cos A}{a} + \frac{\cos B}{b} + 2 \frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca} \]

### Step 1: Apply the Cosine Rule
Using the cosine rule, we know that:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

Substituting these into the equation gives:
\[ 2 \frac{\frac{b^2 + c^2 - a^2}{2bc}}{a} + \frac{\frac{a^2 + c^2 - b^2}{2ac}}{b} + 2 \frac{\frac{a^2 + b^2 - c^2}{2ab}}{c} = \frac{a}{bc} + \frac{b}{ca} \]

### Step 2: Simplify the Left Side
The left side simplifies to:
\[ \frac{b^2 + c^2 - a^2}{ab} + \frac{a^2 + c^2 - b^2}{2bc} + \frac{a^2 + b^2 - c^2}{ab} \]
Combining the terms gives:
\[ \frac{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2)}{2bc} \]

### Step 3: Combine and Rearrange
Combining all terms in the numerator:
\[ = \frac{2a^2 + 2b^2 + 2c^2 - 2c^2}{2bc} = \frac{2a^2 + 2b^2}{2bc} = \frac{a^2 + b^2}{bc} \]

### Step 4: Equate the Two Sides
Now we equate the left side to the right side:
\[ \frac{a^2 + b^2}{bc} = \frac{a}{bc} + \frac{b}{ca} \]
Multiply through by \(bc\):
\[ a^2 + b^2 = a + \frac{b^2}{a} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ a^2 - a - b^2 = 0 \]
This is a quadratic equation in \(a\).

### Step 6: Solve for \(a\)
Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = \frac{1 \pm \sqrt{1 + 4b^2}}{2} \]

### Step 7: Analyze the Triangle
Since we are looking for angle \(A\), we can use the Pythagorean theorem:
\[ a^2 = b^2 + c^2 \]
This indicates that triangle ABC is a right triangle with angle \(A\) being \(90^\circ\).

### Conclusion
Thus, the value of angle \(A\) is:
\[ \boxed{90^\circ} \]