If `5 tan theta =4` , then `frac(5sin theta - 3cos theta )(5 sin theta + 2cos theta )` is equal to

Options

A. `frac(1)(4)`
B. `frac(1)(6)`
C. `frac(1)(3)`
D. `frac(2)(3)`

Answer

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

**Given:**
\[ 5 \tan \theta = 4 \]

### Step 1: Find \(\tan \theta\)

From the equation, we can express \(\tan \theta\) as:
\[ \tan \theta = \frac{4}{5} \]

### Step 2: Construct a right triangle

Using the definition of tangent, we can construct a right triangle where:
- Opposite side (perpendicular) = 4
- Adjacent side (base) = 5

### Step 3: Calculate the hypotenuse

Using the Pythagorean theorem:
\[ h = \sqrt{(4^2) + (5^2)} = \sqrt{16 + 25} = \sqrt{41} \]

### Step 4: Find \(\sin \theta\) and \(\cos \theta\)

Now we can find \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{\sqrt{41}} \]
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{5}{\sqrt{41}} \]

### Step 5: Substitute \(\sin \theta\) and \(\cos \theta\) into the expression

We need to evaluate:
\[ \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} \]

Substituting the values of \(\sin \theta\) and \(\cos \theta\):
\[ = \frac{5 \left(\frac{4}{\sqrt{41}}\right) - 3 \left(\frac{5}{\sqrt{41}}\right)}{5 \left(\frac{4}{\sqrt{41}}\right) + 2 \left(\frac{5}{\sqrt{41}}\right)} \]

### Step 6: Simplify the expression

This simplifies to:
\[ = \frac{\frac{20}{\sqrt{41}} - \frac{15}{\sqrt{41}}}{\frac{20}{\sqrt{41}} + \frac{10}{\sqrt{41}}} \]

Combining the fractions:
\[ = \frac{\frac{20 - 15}{\sqrt{41}}}{\frac{20 + 10}{\sqrt{41}}} \]
\[ = \frac{\frac{5}{\sqrt{41}}}{\frac{30}{\sqrt{41}}} \]

### Step 7: Further simplify

The \(\sqrt{41}\) cancels out:
\[ = \frac{5}{30} = \frac{1}{6} \]

### Final Answer

Thus, the value of the expression is:
\[ \frac{1}{6} \]

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