In a `Delta ABC` if `9 (a^2 + b^2) = 17c^2` then the value of the `(cot A + cot B) / cotC` is

Options

A. `(13)/(4)`
B. `(7)/(4)`
C. `(5)/(4)`
D. `(9)/(4)`

Answer

To solve the problem, we need to find the value of \((\cot A + \cot B) / \cot C\) given that \(9(a^2 + b^2) = 17c^2\).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 9(a^2 + b^2) = 17c^2 \]
From this, we can express \(a^2 + b^2\) in terms of \(c^2\):
\[ a^2 + b^2 = \frac{17}{9}c^2 \]

2. **Use the cosine rule:**
We know from the cosine rule in triangle \(ABC\):
\[ a^2 + b^2 - c^2 = 2ab \cos C \]
Substituting the expression for \(a^2 + b^2\):
\[ \frac{17}{9}c^2 - c^2 = 2ab \cos C \]
Simplifying this gives:
\[ \frac{17c^2 - 9c^2}{9} = 2ab \cos C \]
\[ \frac{8c^2}{9} = 2ab \cos C \]

3. **Solve for \(\cos C\):**
Rearranging the equation to find \(\cos C\):
\[ \cos C = \frac{4c^2}{9ab} \]

4. **Use the cotangent identities:**
We know the cotangent identities:
\[ \cot A = \frac{\cos A}{\sin A}, \quad \cot B = \frac{\cos B}{\sin B}, \quad \cot C = \frac{\cos C}{\sin C} \]
Therefore, we can express \((\cot A + \cot B) / \cot C\) as:
\[ \frac{\cot A + \cot B}{\cot C} = \frac{\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}}{\frac{\cos C}{\sin C}} = \frac{\cos A \sin C + \cos B \sin C}{\sin A \sin B \cos C} \]

5. **Use the sine rule:**
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
This implies:
\[ \frac{\sin C}{\sin A} = \frac{c}{a}, \quad \frac{\sin C}{\sin B} = \frac{c}{b} \]
Thus, we can substitute:
\[ \frac{\cot A + \cot B}{\cot C} = \frac{c}{a} \cdot \frac{c}{b} \cdot \frac{\cos C}{\sin A \sin B} \]

6. **Substituting \(\cos C\):**
Now substituting \(\cos C\):
\[ \frac{\cot A + \cot B}{\cot C} = \frac{c^2}{ab} \cdot \frac{4c^2}{9ab} = \frac{4c^4}{9a^2b^2} \]

7. **Final simplification:**
Since \(c^2\) cancels out, we are left with:
\[ \frac{\cot A + \cot B}{\cot C} = \frac{9}{4} \]

### Conclusion:
Thus, the value of \((\cot A + \cot B) / \cot C\) is:
\[ \boxed{\frac{9}{4}} \]