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 the relationship \(9(a^2 + b^2) = 17c^2\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Use the Cosine Rule**:
The cosine rule states that:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
From the given equation, we can express \(a^2 + b^2\):
\[ a^2 + b^2 = \frac{17}{9}c^2 \]

2. **Substituting into Cosine Rule**:
Substitute \(a^2 + b^2\) into the cosine rule:
\[ \cos C = \frac{\frac{17}{9}c^2 - c^2}{2ab} \]
Simplifying the numerator:
\[ \cos C = \frac{\frac{17}{9}c^2 - \frac{9}{9}c^2}{2ab} = \frac{\frac{8}{9}c^2}{2ab} = \frac{4c^2}{9ab} \]

3. **Express Cotangents**:
Recall the definitions:
\[ \cot A = \frac{\cos A}{\sin A}, \quad \cot B = \frac{\cos B}{\sin B}, \quad \cot C = \frac{\cos C}{\sin C} \]
Therefore, we can express \(\cot A + \cot B\):
\[ \cot A + \cot B = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\cos A \sin B + \sin A \cos B}{\sin A \sin B} = \frac{\sin(A + B)}{\sin A \sin B} \]
Since \(A + B + C = 180^\circ\), we have \(A + B = 180^\circ - C\), thus:
\[ \sin(A + B) = \sin C \]
So,
\[ \cot A + \cot B = \frac{\sin C}{\sin A \sin B} \]

4. **Substituting into the Final Expression**:
Now, we can write:
\[ \frac{\cot A + \cot B}{\cot C} = \frac{\frac{\sin C}{\sin A \sin B}}{\frac{\cos C}{\sin C}} = \frac{\sin^2 C}{\sin A \sin B \cos C} \]

5. **Using the Sine Rule**:
From the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
This gives:
\[ \frac{\sin C}{\sin A} = \frac{c}{a}, \quad \frac{\sin C}{\sin B} = \frac{c}{b} \]
Thus,
\[ \sin A = \frac{a \sin C}{c}, \quad \sin B = \frac{b \sin C}{c} \]
Therefore:
\[ \sin A \sin B = \frac{ab \sin^2 C}{c^2} \]

6. **Substituting Back**:
Substitute \(\sin A \sin B\) into the expression:
\[ \frac{\sin^2 C}{\sin A \sin B \cos C} = \frac{\sin^2 C}{\frac{ab \sin^2 C}{c^2} \cdot \cos C} = \frac{c^2}{ab \cos C} \]

7. **Substituting \(\cos C\)**:
We already found that \(\cos C = \frac{4c^2}{9ab}\), so:
\[ \frac{c^2}{ab \cdot \frac{4c^2}{9ab}} = \frac{c^2 \cdot 9ab}{4c^2} = \frac{9}{4} \]

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