Discussion
If \(\theta = cot^{-1}\sqrt{\dfrac{1 - x}{1 + x}}\), then \(sec^{2}\theta\)
Answer: C
\(\sec^{2}\theta = \) \(\frac{2}{1-x}\).
\(\theta = \cot^{-1}\sqrt{\frac{1 - x}{1 + x}}\) means \(\cot\theta = \sqrt{\frac{1 - x}{1 + x}}\), so \(\tan^{2}\theta = \frac{1 + x}{1 - x}\).
\[ \begin{aligned} \sec^{2}\theta &= 1 + \tan^{2}\theta \\ &= \frac{(1 - x) + (1 + x)}{1 - x} = \frac{2}{1 - x} \end{aligned} \]
\(\frac{1 - x}{2}\) is \(\cos^{2}\theta\), the reciprocal of the answer, and \(\frac{1 + x}{2}\) is \(\sin^{2}\theta\); \(x\) and \(2x\) do not come from the identity at all.
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