Question

If $\sin \theta = \frac{1}{5}$ and the angle $\theta$ is in the second quadrant, then $\sec \theta$ is equal to:

• A. $\frac{5}{2\sqrt{6}}$
• B. $\frac{-2\sqrt{6}}{5}$
• C. $\frac{2\sqrt{6}}{5}$
• D. $\frac{\sqrt{6}}{5}$
• E. $\frac{-5}{2\sqrt{6}}$

Hint:

"All Silver Tea Cups" helps remember which functions are positive in each quadrant (S for Sine is positive in the 2nd).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are given the value of \( \sin \theta \) and the quadrant in which \( \theta \) lies. We need to find the value of \( \sec \theta \). This requires using the Pythagorean identity to find \( \cos \theta \) and then the reciprocal identity for \( \sec \theta \), paying close attention to the sign conventions in the specified quadrant.
Step 2: Key Formula or Approach:
1. Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
2. Reciprocal identity: \( \sec \theta = \frac{1}{\cos \theta} \)
3. Quadrant rules: In the second quadrant, sine is positive, while cosine and secant are negative.
Step 3: Detailed Explanation:
We are given \( \sin \theta = \frac{1}{5} \).
Using the Pythagorean identity, we can find \( \cos^2 \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \] \[ \cos^2 \theta = 1 - \left(\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \] Now, we find \( \cos \theta \) by taking the square root:
\[ \cos \theta = \pm \sqrt{\frac{24}{25}} = \pm \frac{\sqrt{24}}{5} = \pm \frac{\sqrt{4 \cdot 6}}{5} = \pm \frac{2\sqrt{6}}{5} \] The problem states that \( \theta \) is in the second quadrant. In the second quadrant, the cosine function is negative. Therefore, we choose the negative value:
\[ \cos \theta = -\frac{2\sqrt{6}}{5} \] Finally, we find \( \sec \theta \) using the reciprocal identity:
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{2\sqrt{6}}{5}} = -\frac{5}{2\sqrt{6}} \] Step 4: Final Answer:
The value of \( \sec \theta \) is \( -\frac{5}{2\sqrt{6}} \).