If `5tantheta =4," then " (5sintheta-3costheta)/(5sintheta+2costheta)` is equal to

Options

A. `0`
B. `-6`
C. `1/6`
D. `6`

Explanation

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

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

**Step 1: Find the value of \(\tan \theta\)**
From the equation, we can isolate \(\tan \theta\):
\[ \tan \theta = \frac{4}{5} \]

**Step 2: Express \(\sin \theta\) and \(\cos \theta\) in terms of \(\tan \theta\)**
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Let \(\sin \theta = 4k\) and \(\cos \theta = 5k\) for some \(k\). Then:
\[ \tan \theta = \frac{4k}{5k} = \frac{4}{5} \]

**Step 3: Use the Pythagorean identity**
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ (4k)^2 + (5k)^2 = 1 \]
\[ 16k^2 + 25k^2 = 1 \]
\[ 41k^2 = 1 \]
\[ k^2 = \frac{1}{41} \]
\[ k = \frac{1}{\sqrt{41}} \]

**Step 4: Find \(\sin \theta\) and \(\cos \theta\)**
Now substituting back for \(k\):
\[ \sin \theta = 4k = \frac{4}{\sqrt{41}}, \quad \cos \theta = 5k = \frac{5}{\sqrt{41}} \]

**Step 5: Substitute \(\sin \theta\) and \(\cos \theta\) into the expression**
We need to find:
\[ \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} \]
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:
\[ = \frac{\frac{20}{\sqrt{41}} - \frac{15}{\sqrt{41}}}{\frac{20}{\sqrt{41}} + \frac{10}{\sqrt{41}}} \]
\[ = \frac{\frac{5}{\sqrt{41}}}{\frac{30}{\sqrt{41}}} \]
\[ = \frac{5}{30} = \frac{1}{6} \]

**Final Answer:**
\[ \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = \frac{1}{6} \]

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