if sin(θ)=2/3 and θ is in quadrant 2 find sec(θ)

To find \( \sec(\theta) \), we first need to recall that:

\[
\sec(\theta) = \frac{1}{\cos(\theta)}
\]

We know that \( \sin(\theta) = \frac{2}{3} \), and since \( \theta \) is in the second quadrant, cosine is negative in this quadrant.

1. Start by using the Pythagorean identity:
   \[
   \sin^2(\theta) + \cos^2(\theta) = 1
   \]
   Substitute \( \sin(\theta) = \frac{2}{3} \):
   \[
   \left(\frac{2}{3}\right)^2 + \cos^2(\theta) = 1
   \]
   \[
   \frac{4}{9} + \cos^2(\theta) = 1
   \]
   \[
   \cos^2(\theta) = 1 - \frac{4}{9} = \frac{5}{9}
   \]
   \[
   \cos(\theta) = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}
   \]
   (The negative sign is because \( \theta \) is in the second quadrant, where cosine is negative.)

2. Now calculate \( \sec(\theta) \):
   \[
   \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{\sqrt{5}}{3}} = -\frac{3}{\sqrt{5}}
   \]

3. To rationalize the denominator:
   \[
   \sec(\theta) = -\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = -\frac{3\sqrt{5}}{5}
   \]

Thus, \( \sec(\theta) = -\frac{3\sqrt{5}}{5} \).

Would you like further details on the steps, or do you have any questions?

Here are five related questions to expand on this topic:
1. How do the signs of trigonometric functions vary in each quadrant?
2. What is the relationship between sine and cosine in different quadrants?
3. How can you find other trigonometric functions given one value in a specific quadrant?
4. How do you rationalize denominators when dealing with square roots?
5. Can you prove the Pythagorean identity for any angle?

**Tip**: When solving trigonometric problems, always pay attention to the signs of the functions based on the quadrant the angle is in.

If sin(θ)=2/3 and θ is in quadrant 2, find sec(θ).

Learn how to find sec(θ) when sin(θ) = 2/3 and θ is in the second quadrant. This step-by-step solution uses trigonometric identities, including the Pythagorean identity and sign conventions for different quadrants. Suitable for Grades 10-12.

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