If `tantheta=(11)/(13)`, then find `(5sintheta-3costheta)/(5sintheta+2costheta)`

Options

A. `(81)/(16)`
B. `(9)/(4)`
C. `(16)/(81)`
D. `(4)/(9)`

Answer

To solve the problem where \( \tan \theta = \frac{11}{13} \) and we need to find \( \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} \), we can follow these steps:

### Step 1: Express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan \theta \)
Given \( \tan \theta = \frac{11}{13} \), we can use the definitions of sine and cosine:
\[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} \quad \text{and} \quad \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} \]
Calculating \( 1 + \tan^2 \theta \):
\[ 1 + \tan^2 \theta = 1 + \left(\frac{11}{13}\right)^2 = 1 + \frac{121}{169} = \frac{169 + 121}{169} = \frac{290}{169} \]
Thus,
\[ \sqrt{1 + \tan^2 \theta} = \sqrt{\frac{290}{169}} = \frac{\sqrt{290}}{13} \]
Now we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{\frac{11}{13}}{\frac{\sqrt{290}}{13}} = \frac{11}{\sqrt{290}} \quad \text{and} \quad \cos \theta = \frac{1}{\frac{\sqrt{290}}{13}} = \frac{13}{\sqrt{290}} \]

### Step 2: Substitute \( \sin \theta \) and \( \cos \theta \) into the expression
Now substitute \( \sin \theta \) and \( \cos \theta \) into the expression:
\[ \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = \frac{5 \left(\frac{11}{\sqrt{290}}\right) - 3 \left(\frac{13}{\sqrt{290}}\right)}{5 \left(\frac{11}{\sqrt{290}}\right) + 2 \left(\frac{13}{\sqrt{290}}\right)} \]
This simplifies to:
\[ = \frac{\frac{55 - 39}{\sqrt{290}}}{\frac{55 + 26}{\sqrt{290}}} = \frac{16/\sqrt{290}}{81/\sqrt{290}} \]

### Step 3: Simplify the expression
The \( \sqrt{290} \) cancels out:
\[ = \frac{16}{81} \]

### Final Answer
Thus, the final answer is:
\[ \frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = \frac{16}{81} \]