If `tantheta=(5)/(9)` , then what is the value of `(5sintheta+9costheta)/(5sintheta-9costheta)` ?

Options

A. `(17)/(12)`
B. `(-53)/(28)`
C. `(-27)/(25)`
D. `(31)/(23)`

Answer

To solve the problem, we start with the given information:

1. **Given:** \( \tan \theta = \frac{5}{9} \)

We need to find the value of:

\[ \frac{5 \sin \theta + 9 \cos \theta}{5 \sin \theta - 9 \cos \theta} \]

### Step 1: Express \(\sin \theta\) and \(\cos \theta\) in terms of \(\tan \theta\)

From the definition of tangent, we know:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Given \( \tan \theta = \frac{5}{9} \), we can let:

\[ \sin \theta = 5k \quad \text{and} \quad \cos \theta = 9k \]

for some \( k \).

### Step 2: Find \( k \) using the Pythagorean identity

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ (5k)^2 + (9k)^2 = 1 \]

This simplifies to:

\[ 25k^2 + 81k^2 = 1 \]
\[ 106k^2 = 1 \]
\[ k^2 = \frac{1}{106} \]
\[ k = \frac{1}{\sqrt{106}} \]

### Step 3: Substitute \( k \) back to find \(\sin \theta\) and \(\cos \theta\)

Now substituting \( k \) back:

\[ \sin \theta = 5k = \frac{5}{\sqrt{106}}, \quad \cos \theta = 9k = \frac{9}{\sqrt{106}} \]

### Step 4: Substitute \(\sin \theta\) and \(\cos \theta\) into the expression

Now substitute these values into the expression:

\[ \frac{5 \sin \theta + 9 \cos \theta}{5 \sin \theta - 9 \cos \theta} = \frac{5 \left(\frac{5}{\sqrt{106}}\right) + 9 \left(\frac{9}{\sqrt{106}}\right)}{5 \left(\frac{5}{\sqrt{106}}\right) - 9 \left(\frac{9}{\sqrt{106}}\right)} \]

This simplifies to:

\[ = \frac{\frac{25}{\sqrt{106}} + \frac{81}{\sqrt{106}}}{\frac{25}{\sqrt{106}} - \frac{81}{\sqrt{106}}} \]

### Step 5: Simplify the expression

Combining the fractions:

\[ = \frac{\frac{106}{\sqrt{106}}}{\frac{-56}{\sqrt{106}}} \]

This simplifies to:

\[ = \frac{106}{-56} = -\frac{53}{28} \]

### Final Answer

Thus, the value of \( \frac{5 \sin \theta + 9 \cos \theta}{5 \sin \theta - 9 \cos \theta} \) is:

\[ -\frac{53}{28} \]