What is \cot^{2}(\sec^{-1}2)+\tan^{2}(\text{cosec}^{-1}3) equal to?
- A. 11/12
- B. 11/24 ✓
- C. 7/24
- D. 1/24
Correct Answer: B. 11/24
Explanation
Let \sec^{-1}2 = A, then \sec A = 2, which implies \cos A = \frac{1}{2}, so A = \frac{\pi}{3}. Thus, \cot^2(\frac{\pi}{3}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}. Let \text{cosec}^{-1}3 = B, then \text{cosec} B = 3, which implies \sin B = \frac{1}{3}. We know \tan^2 B = \frac{\sin^2 B}{\cos^2 B} = \frac{\sin^2 B}{1 - \sin^2 B} = \frac{1/9}{1 - 1/9} = \frac{1/9}{8/9} = \frac{1}{8}. The sum is \frac{1}{3} + \frac{1}{8} = \frac{11}{24}.