Question

If \( \sin \alpha = \frac{12}{13} \), and \( \frac{\pi}{6} \leq \alpha \leq \frac{3\pi}{2} \), find \( \tan \alpha \).

• A. \( \frac{5}{12} \)
• B. \( \frac{5}{13} \)
• C. \( \frac{12}{5} \)
• D. \( \frac{13}{5} \)

Hint:

Use the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \) to find missing trigonometric values. Also, remember to account for the signs of trigonometric functions in different quadrants.

Answer 1

The Correct Option is C

We are given that \( \sin \alpha = \frac{12}{13} \) and \( \frac{\pi}{6} \leq \alpha \leq \frac{3\pi}{2} \), which means \( \alpha \) is in the second or third quadrant (since \( \frac{\pi}{6} \leq \alpha \leq \frac{3\pi}{2} \)).

1. Step 1: Use the Pythagorean identity From the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we can find \( \cos \alpha \). Since \( \sin \alpha = \frac{12}{13} \), we substitute this value into the identity: \[ \left( \frac{12}{13} \right)^2 + \cos^2 \alpha = 1 \] \[ \frac{144}{169} + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] \[ \cos \alpha = \pm \frac{5}{13} \]

2. Step 2: Determine the sign of \( \cos \alpha \) Since \( \alpha \) is in the second or third quadrant, and sine is positive in the second quadrant and cosine is negative, we choose \( \cos \alpha = -\frac{5}{13} \) (since \( \alpha \) must be in the second quadrant).

3. Step 3: Find \( \tan \alpha \) Now that we have \( \sin \alpha = \frac{12}{13} \) and \( \cos \alpha = -\frac{5}{13} \), we can use the formula for the tangent: \[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \] Thus, \( \tan \alpha = \frac{12}{5} \) (taking the absolute value since tangent is positive in the third quadrant).