If `5 cos theta = 6 , sin theta `, evalutate :
`(12 sin theta - 3cos theta)/(12 sin theta + 3 cos theta)`

Answer

To solve the problem, we need to evaluate the expression \((12 \sin \theta - 3 \cos \theta) / (12 \sin \theta + 3 \cos \theta)\) given that \(5 \cos \theta = 6 \sin \theta\).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ 5 \cos \theta = 6 \sin \theta \]

2. **Rearrange the equation**:
\[ \frac{\sin \theta}{\cos \theta} = \frac{5}{6} \]
This implies:
\[ \tan \theta = \frac{5}{6} \]

3. **Substitute \(\tan \theta\) into the expression**:
We need to evaluate:
\[ \frac{12 \sin \theta - 3 \cos \theta}{12 \sin \theta + 3 \cos \theta} \]
To do this, we can multiply the numerator and the denominator by \(\cos \theta\):
\[ = \frac{12 \sin \theta \cdot \cos \theta - 3 \cos^2 \theta}{12 \sin \theta \cdot \cos \theta + 3 \cos^2 \theta} \]

4. **Express \(\sin \theta\) and \(\cos \theta\) in terms of \(\tan \theta\)**:
Since \(\tan \theta = \frac{5}{6}\), we can express \(\sin \theta\) and \(\cos \theta\) as follows:
Let \(k = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}\).
Then:
\[ \sin \theta = \frac{5}{\sqrt{61}}, \quad \cos \theta = \frac{6}{\sqrt{61}} \]

5. **Substitute \(\sin \theta\) and \(\cos \theta\) back into the expression**:
Now substituting these values into the expression:
\[ \frac{12 \cdot \frac{5}{\sqrt{61}} - 3 \cdot \frac{6}{\sqrt{61}}}{12 \cdot \frac{5}{\sqrt{61}} + 3 \cdot \frac{6}{\sqrt{61}}} \]
This simplifies to:
\[ = \frac{\frac{60}{\sqrt{61}} - \frac{18}{\sqrt{61}}}{\frac{60}{\sqrt{61}} + \frac{18}{\sqrt{61}}} \]
\[ = \frac{\frac{42}{\sqrt{61}}}{\frac{78}{\sqrt{61}}} \]
\[ = \frac{42}{78} \]

6. **Simplify the fraction**:
\[ = \frac{7}{13} \]

### Final Answer:
Thus, the value of \(\frac{12 \sin \theta - 3 \cos \theta}{12 \sin \theta + 3 \cos \theta}\) is \(\frac{7}{13}\).