If f(\alpha)=\sqrt{\sec^{2}\alpha-1}, then what is \frac{f(\alpha)+f(\beta)}{1-f(\alpha)f(\beta)} equal to ?
- A. f(\alpha-\beta)
- B. f(\alpha+\beta) ✓
- C. f(\alpha)f(\beta)
- D. f(\alpha\beta)
Correct Answer: B. f(\alpha+\beta)
Explanation
Given f(\alpha) = \sqrt{\sec^2\alpha - 1} = \tan\alpha. Substituting this into the expression, we get \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}, which is the trigonometric identity for \tan(\alpha+\beta). Thus, it is equal to f(\alpha+\beta).