If 2 \tan \theta = \sec^2 \theta - 2, where 0 < \theta < \pi/2, then what is \cot \theta equal to?
- A. \sqrt{2} - 1 ✓
- B. \sqrt{2} + 1
- C. \sqrt{3} - 1
- D. \sqrt{3} + 2
Correct Answer: A. \sqrt{2} - 1
Explanation
Substituting \sec^2 \theta = 1 + \tan^2 \theta, we get 2\tan \theta = \tan^2 \theta - 1. Rearranging gives \tan^2 \theta - 2\tan \theta - 1 = 0. Solving for \tan \theta yields 1 \pm \sqrt{2}. Since 0 < \theta < \pi/2, \tan \theta = \sqrt{2} + 1. Thus, \cot \theta = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1.
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