Discussion
If \(sec^{2}\theta + tan^{2}\theta = 7, 0 \lt \theta \lt \frac{\pi}{2}\) , then \(tan2\theta\) is equal to
Answer: D
\(\tan 2\theta\) is \(-\sqrt{3}\).
Replace \(\sec^{2}\theta\) by \(1 + \tan^{2}\theta\).
\[ \begin{aligned} 1 + 2\tan^{2}\theta &= 7 \Rightarrow \tan^{2}\theta = 3 \\ \tan\theta &= \sqrt{3},\ \theta = \tfrac{\pi}{3} \\ \tan 2\theta &= \tan\tfrac{2\pi}{3} = -\sqrt{3} \end{aligned} \]
The positive root is taken because \(\theta\) is in the first quadrant. The double angle \(\frac{2\pi}{3}\) is in the second quadrant, where tangent is negative, so \(\sqrt{3}\) is the sign trap. The formula \(\frac{2\tan\theta}{1 - \tan^{2}\theta} = \frac{2\sqrt{3}}{-2}\) gives the same result.
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