Question

If \( \sec(\alpha + \beta) = \frac{\sqrt{7}}{\sqrt{3}} \), then \( \sin(\alpha + \beta) + \tan(\alpha + \beta) \) is equal to:

• A. \( \frac{\sqrt{3} + \sqrt{7}}{\sqrt{21}} \)
• B. \( \frac{2}{\sqrt{21}} \)
• C. \( \frac{2(\sqrt{3} + \sqrt{7})}{\sqrt{21}} \)
• D. \( \frac{\sqrt{7}}{\sqrt{3}} \)
• E. \( \frac{\sqrt{3}}{\sqrt{7}} \)

Hint:

When dealing with trigonometric identities, use \( \sec(\theta) = \frac{1}{\cos(\theta)} \) to find \( \cos(\theta) \), and use \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) to find \( \tan(\theta) \).

Answer 1

The Correct Option is C

We are given that \( \sec(\alpha + \beta) = \frac{\sqrt{7}}{\sqrt{3}} \), and we are asked to find \( \sin(\alpha + \beta) + \tan(\alpha + \beta) \). 
Step 1: From the given, \( \sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} \), so we can write: \[ \cos(\alpha + \beta) = \frac{\sqrt{3}}{\sqrt{7}}. \] Step 2: Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \sin(\alpha + \beta) \): \[ \sin^2(\alpha + \beta) = 1 - \cos^2(\alpha + \beta) = 1 - \left( \frac{\sqrt{3}}{\sqrt{7}} \right)^2 = 1 - \frac{3}{7} = \frac{4}{7}. \] Thus, \[ \sin(\alpha + \beta) = \frac{2}{\sqrt{7}}. \] Step 3: Now, we can calculate \( \tan(\alpha + \beta) \) using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \): \[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{2}{\sqrt{7}}}{\frac{\sqrt{3}}{\sqrt{7}}} = \frac{2}{\sqrt{3}}. \] Step 4: Finally, we calculate the sum \( \sin(\alpha + \beta) + \tan(\alpha + \beta) \): \[ \sin(\alpha + \beta) + \tan(\alpha + \beta) = \frac{2}{\sqrt{7}} + \frac{2}{\sqrt{3}}. \] To combine these, we need a common denominator: \[ \frac{2}{\sqrt{7}} + \frac{2}{\sqrt{3}} = \frac{2\sqrt{3} + 2\sqrt{7}}{\sqrt{21}} = \frac{2(\sqrt{3} + \sqrt{7})}{\sqrt{21}}. \] Thus, the correct answer is option (C).