What is \sqrt{15+\cot^{2}(\frac{\pi}{4}-2 \cot^{-1}3)} equal to ?
- A. 1
- B. 7
- C. 8 ✓
- D. 16
Correct Answer: C. 8
Explanation
Let \theta = \cot^{-1} 3 \implies \tan \theta = \frac{1}{3}. Using the double angle formula, \tan 2\theta = \frac{2(1/3)}{1-(1/3)^2} = \frac{3}{4}. Then, \tan(\frac{\pi}{4}-2\theta) = \frac{1-\tan 2\theta}{1+\tan 2\theta} = \frac{1-3/4}{1+3/4} = \frac{1}{7}. This means \cot(\frac{\pi}{4}-2\theta) = 7. The given expression becomes \sqrt{15 + 7^2} = \sqrt{15 + 49} = \sqrt{64} = 8.