if `tantheta=(4)/(3)`, then the value of `(3sintheta+2costheta)/(3sintheta-2costheta)` is

Answer

To solve the problem, we need to find the value of \(\frac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta}\) given that \(\tan \theta = \frac{4}{3}\).

### Step-by-Step Solution:

1. **Start with the given information**:
\[ \tan \theta = \frac{4}{3} \]

2. **Express \(\sin \theta\) and \(\cos \theta\)**:
From the definition of tangent, we know:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Therefore, we can let:
\[ \sin \theta = 4k \quad \text{and} \quad \cos \theta = 3k \]
for some positive value \(k\).

3. **Find the value of \(k\)** using the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting our expressions for \(\sin \theta\) and \(\cos \theta\):
\[ (4k)^2 + (3k)^2 = 1 \]
\[ 16k^2 + 9k^2 = 1 \]
\[ 25k^2 = 1 \]
\[ k^2 = \frac{1}{25} \quad \Rightarrow \quad k = \frac{1}{5} \]

4. **Calculate \(\sin \theta\) and \(\cos \theta\)**:
\[ \sin \theta = 4k = 4 \times \frac{1}{5} = \frac{4}{5} \]
\[ \cos \theta = 3k = 3 \times \frac{1}{5} = \frac{3}{5} \]

5. **Substitute \(\sin \theta\) and \(\cos \theta\) into the expression**:
Now we substitute into the original expression:
\[ \frac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta} = \frac{3 \left(\frac{4}{5}\right) + 2 \left(\frac{3}{5}\right)}{3 \left(\frac{4}{5}\right) - 2 \left(\frac{3}{5}\right)} \]

6. **Simplify the numerator**:
\[ = \frac{\frac{12}{5} + \frac{6}{5}}{\frac{12}{5} - \frac{6}{5}} = \frac{\frac{18}{5}}{\frac{6}{5}} \]

7. **Simplify the fraction**:
\[ = \frac{18}{5} \times \frac{5}{6} = \frac{18}{6} = 3 \]

### Final Answer:
Thus, the value of \(\frac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta}\) is \(3\).