What is \cot^{-1}9+\csc^{-1}\left(\frac{\sqrt{41}}{4}\right) equal to?
- A. \frac{\pi}{4} ✓
- B. \frac{\pi}{3}
- C. \frac{\pi}{2}
- D. \pi
Correct Answer: A. \frac{\pi}{4}
Explanation
Let \alpha = \cot^{-1}9, so \tan\alpha = \frac{1}{9}. Let \beta = \csc^{-1}\left(\frac{\sqrt{41}}{4}\right), giving a right triangle with opposite 4 and hypotenuse \sqrt{41}, so adjacent is 5; thus \tan\beta = \frac{4}{5}. We find \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{1/9 + 4/5}{1 - 4/45} = \frac{41/45}{41/45} = 1. Therefore, \alpha + \beta = \frac{\pi}{4}.