If `(cosec^(2)theta-1)=(144)/(25)andtheta` is acute , then what is the value of `sqrt((cottheta+tantheta))`?

Options

A. `(13)/(5)`
B. `(60)/(13)`
C. `(2sqrt(15))/(13)`
D. `(13)/(2sqrt(15))`

Answer

To solve the problem, we need to find the value of \( \sqrt{\cot \theta + \tan \theta} \) given that \( \csc^2 \theta - 1 = \frac{144}{25} \) and \( \theta \) is acute.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \csc^2 \theta - 1 = \frac{144}{25} \]

2. **Rearranging the equation to find \( \csc^2 \theta \):**
\[ \csc^2 \theta = \frac{144}{25} + 1 \]
To add \( 1 \), we convert it to a fraction:
\[ 1 = \frac{25}{25} \]
Therefore,
\[ \csc^2 \theta = \frac{144}{25} + \frac{25}{25} = \frac{169}{25} \]

3. **Finding \( \csc \theta \):**
\[ \csc \theta = \sqrt{\csc^2 \theta} = \sqrt{\frac{169}{25}} = \frac{13}{5} \]

4. **Using the definition of \( \csc \theta \):**
Since \( \csc \theta = \frac{\text{hypotenuse}}{\text{opposite}} \), we can assume:
- Hypotenuse = 13
- Opposite = 5

5. **Finding the base using Pythagorean theorem:**
\[ \text{Base}^2 = \text{Hypotenuse}^2 - \text{Opposite}^2 \]
\[ \text{Base}^2 = 13^2 - 5^2 = 169 - 25 = 144 \]
\[ \text{Base} = \sqrt{144} = 12 \]

6. **Calculating \( \cot \theta \) and \( \tan \theta \):**
- \( \cot \theta = \frac{\text{Base}}{\text{Opposite}} = \frac{12}{5} \)
- \( \tan \theta = \frac{\text{Opposite}}{\text{Base}} = \frac{5}{12} \)

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

8. **Taking the square root:**
\[ \sqrt{\cot \theta + \tan \theta} = \sqrt{\frac{169}{60}} = \frac{\sqrt{169}}{\sqrt{60}} = \frac{13}{\sqrt{60}} \]

9. **Simplifying \( \sqrt{60} \):**
\[ \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \]
Thus,
\[ \sqrt{\cot \theta + \tan \theta} = \frac{13}{2\sqrt{15}} \]

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