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We are given the following problem:

- \(\sin \theta = \frac{3}{5}\)
- \(\theta\) is in the second quadrant (\( \theta \in II \) cuadrante)
- The goal is to find \( \sec \theta \).

### Step 1: Using the Pythagorean identity to find \( \cos \theta \)

Since we know \(\sin \theta = \frac{3}{5}\), we can use the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substitute \(\sin \theta = \frac{3}{5}\):
\[
\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1
\]
\[
\frac{9}{25} + \cos^2 \theta = 1
\]
\[
\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}
\]
\[
\cos \theta = \pm \frac{4}{5}
\]

### Step 2: Determine the sign of \( \cos \theta \)

In the second quadrant, cosine is negative. Therefore, \( \cos \theta = -\frac{4}{5} \).

### Step 3: Find \( \sec \theta \)

The secant function is the reciprocal of the cosine function:
\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}
\]

### Answer:
The value of \( \sec \theta \) is \( -\frac{5}{4} \), which corresponds to option (d).

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Do you want more details or have any questions about this? 

Here are 5 related questions to expand on this concept:
1. How would the value of \( \sec \theta \) change if \( \theta \) were in the third quadrant?
2. What is the relationship between \( \sec \theta \) and \( \cos \theta \)?
3. Can you find \( \tan \theta \) given the same information?
4. How would you use the unit circle to confirm the sign of \( \cos \theta \) in different quadrants?
5. What is the reciprocal of \( \sin \theta \), and how would you find it?

**Tip**: Remember, in the second quadrant, sine is positive and cosine is negative!

Si sin θ = 3/5 y θ ∈ II cuadrante, el valor de sec θ es: (a) 4/3 (b) 5/4 (c) -4/5 (d) -5/4

Learn how to find the value of sec θ when sin θ = 3/5 and θ is in the second quadrant. This trigonometry problem involves using the Pythagorean identity and reciprocal functions. Suitable for high school students in Grades 10-12.

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