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 equation given in the problem:

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

we will use the cosine rule and properties of triangles.

### Step 1: Use the Cosine Rule
The cosine rule states that:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 2: Substitute Cosine Values
Substituting the values of \(\cos A\), \(\cos B\), and \(\cos C\) into the equation:

\[ 2 \frac{\frac{b^2 + c^2 - a^2}{2bc}}{a} + \frac{\frac{a^2 + c^2 - b^2}{2ac}}{b} + 2 \frac{\frac{a^2 + b^2 - c^2}{2ab}}{c} \]

This simplifies to:

\[ \frac{b^2 + c^2 - a^2}{a \cdot bc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{2(a^2 + b^2 - c^2)}{2abc} \]

### Step 3: Combine Terms
Now combine the terms on the left-hand side:

\[ \frac{b^2 + c^2 - a^2 + a^2 + c^2 - b^2 + 2(a^2 + b^2 - c^2)}{2abc} \]

This simplifies to:

\[ \frac{3b^2 + a^2 + c^2 - 2c^2}{2abc} = \frac{3b^2 + a^2 - c^2}{2abc} \]

### Step 4: Right-Hand Side
Now, simplify the right-hand side:

\[ \frac{a}{bc} + \frac{b}{ca} = \frac{a^2 + b^2}{abc} \]

### Step 5: Set Both Sides Equal
Now we equate both sides:

\[ \frac{3b^2 + a^2 - c^2}{2abc} = \frac{a^2 + b^2}{abc} \]

### Step 6: Cross Multiply
Cross-multiplying gives:

\[ 3b^2 + a^2 - c^2 = 2(a^2 + b^2) \]

### Step 7: Rearranging the Equation
Rearranging gives:

\[ 3b^2 + a^2 - c^2 = 2a^2 + 2b^2 \]
\[ b^2 - a^2 - c^2 = 0 \]

### Step 8: Recognizing a Right Triangle
This implies:

\[ b^2 + c^2 = a^2 \]

This is the Pythagorean theorem, which indicates that triangle ABC is a right triangle with angle A being the right angle.

### Conclusion
Thus, the value of angle A is:

\[ \boxed{90^\circ} \]