If `cos alpha=(5)/(13) and 0 lt alpha lt 90^(@)`, then the value of `cot alpha+"cosec "alpha` is

Options

A. 2
B. 3.5
C. 4
D. 1.5

Answer

To solve the problem, we need to find the value of \( \cot \alpha + \csc \alpha \) given that \( \cos \alpha = \frac{5}{13} \) and \( 0 < \alpha < 90^\circ \).

### Step-by-Step Solution:

1. **Identify the Triangle Sides**:
Given \( \cos \alpha = \frac{5}{13} \), we can interpret this in the context of a right triangle where:
- The adjacent side (base) = 5
- The hypotenuse = 13

2. **Use Pythagorean Theorem to Find the Opposite Side**:
According to the Pythagorean theorem:
\[ \text{hypotenuse}^2 = \text{base}^2 + \text{perpendicular}^2 \]
Plugging in the values:
\[ 13^2 = 5^2 + \text{perpendicular}^2 \]
\[ 169 = 25 + \text{perpendicular}^2 \]
\[ \text{perpendicular}^2 = 169 - 25 = 144 \]
Taking the square root:
\[ \text{perpendicular} = \sqrt{144} = 12 \]

3. **Calculate \( \cot \alpha \) and \( \csc \alpha \)**:
- \( \cot \alpha = \frac{\text{base}}{\text{perpendicular}} = \frac{5}{12} \)
- \( \csc \alpha = \frac{\text{hypotenuse}}{\text{perpendicular}} = \frac{13}{12} \)

4. **Add \( \cot \alpha \) and \( \csc \alpha \)**:
\[ \cot \alpha + \csc \alpha = \frac{5}{12} + \frac{13}{12} = \frac{5 + 13}{12} = \frac{18}{12} \]
Simplifying:
\[ \frac{18}{12} = \frac{3}{2} = 1.5 \]

### Final Answer:
Thus, the value of \( \cot \alpha + \csc \alpha \) is \( 1.5 \).

---