If 2 \tan \theta = \sec^2 \theta - 2, where 0 < \theta < \pi/2, then what is \cot \theta equal to?

  1. A. \sqrt{2} - 1
  2. B. \sqrt{2} + 1
  3. C. \sqrt{3} - 1
  4. 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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