Question

If \( \sin \alpha = \frac{12}{13} \), where \( \frac{\pi}{2}<\alpha<\frac{3\pi}{2} \), then the value of \( \tan \alpha \) is equal to

• A. \( \frac{5}{12} \)
• B. \( \frac{13}{5} \)
• C. \( -\frac{12}{5} \)
• D. \( -\frac{13}{5} \)
• E. \( -\frac{1}{12} \)

Hint:

Always check quadrant to assign correct sign for trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires finding the value of a trigonometric ratio given another, along with the quadrant in which the angle lies. The quadrant information is crucial for determining the correct sign of the result.
Step 2: Key Formula or Approach:
1. Use the Pythagorean identity \(\sin^2\alpha + \cos^2\alpha = 1\) to find the value of \(\cos\alpha\). 2. Determine the sign of \(\cos\alpha\) and \(\tan\alpha\) based on the given interval for \(\alpha\). 3. Calculate \(\tan\alpha = \frac{\sin\alpha}{\cos\alpha}\). Alternatively, we can form a right-angled triangle to find the magnitudes of the sides and use the quadrant to determine the sign.
Step 3: Detailed Explanation:
1. Determine the Quadrant: We are given \(\frac{\pi}{2}<\alpha<\frac{3\pi}{2}\), which covers Quadrant II and Quadrant III. We are also given that \(\sin\alpha = \frac{12}{13}\), which is a positive value. The sine function is positive only in Quadrants I and II. The intersection of these two conditions is Quadrant II. In Quadrant II, cosine is negative and tangent is negative. 2. Find the value of \(\cos\alpha\): Using the identity \(\cos^2\alpha = 1 - \sin^2\alpha\): \[ \cos^2\alpha = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \] Taking the square root: \[ \cos\alpha = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \] Since \(\alpha\) is in Quadrant II, \(\cos\alpha\) must be negative. \[ \cos\alpha = -\frac{5}{13} \] 3. Calculate \(\tan\alpha\): \[ \tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{-5/13} = -\frac{12}{5} \] Step 4: Final Answer:
The value of tan \(\alpha\) is \(-\frac{12}{5}\).