If `sectheta=(13)/(12)andtheta` is acute , theh what is the value of `(sqrt(cottheta+tantheta))`?

Options

A. `(13)/(2sqrt(15))`
B. `(12)/(2sqrt(13))`
C. `(13)/(2sqrt(5))`
D. `(2)/(13)`

Answer

To solve the problem, we need to find the value of \(\sqrt{\cot \theta + \tan \theta}\) given that \(\sec \theta = \frac{13}{12}\) and \(\theta\) is an acute angle.

### Step-by-Step Solution:

1. **Understanding Secant**:
\[ \sec \theta = \frac{1}{\cos \theta} \]
Given \(\sec \theta = \frac{13}{12}\), we can find \(\cos \theta\):
\[ \cos \theta = \frac{12}{13} \]

2. **Finding the Sine**:
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{12}{13}\right)^2 \]
\[ = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \]
Thus,
\[ \sin \theta = \sqrt{\frac{25}{169}} = \frac{5}{13} \]

3. **Finding Cotangent and Tangent**:
- **Cotangent**:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \]
- **Tangent**:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

4. **Calculating \(\cot \theta + \tan \theta\)**:
\[ \cot \theta + \tan \theta = \frac{12}{5} + \frac{5}{12} \]
To add these fractions, we need a common denominator:
\[ = \frac{12 \times 12 + 5 \times 5}{5 \times 12} = \frac{144 + 25}{60} = \frac{169}{60} \]

5. **Finding the Square Root**:
Now, we take the square root:
\[ \sqrt{\cot \theta + \tan \theta} = \sqrt{\frac{169}{60}} = \frac{\sqrt{169}}{\sqrt{60}} = \frac{13}{\sqrt{60}} \]

6. **Simplifying \(\sqrt{60}\)**:
We can simplify \(\sqrt{60}\):
\[ \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \]
Therefore,
\[ \sqrt{\cot \theta + \tan \theta} = \frac{13}{2\sqrt{15}} \]

### Final Answer:
\[ \sqrt{\cot \theta + \tan \theta} = \frac{13}{2\sqrt{15}} \]