Discussion
If \(z = 1 + i\ tan\,\theta\) , where \(\pi \lt \theta \lt \frac{3\pi}{2}\) , then \(|z|\) is equal to
Answer: D
\(|z|\) equals \(-\sec\theta\).
For \(z = 1 + i\tan\theta\):
\[ \begin{aligned} |z| &= \sqrt{1 + \tan^{2}\theta} \\ &= \sqrt{\sec^{2}\theta} = |\sec\theta| \end{aligned} \]
For \(\pi \lt \theta \lt \frac{3\pi}{2}\), \(\theta\) is in the third quadrant, where \(\cos\theta\) and so \(\sec\theta\) are negative. A modulus must be positive, so \(|\sec\theta| = -\sec\theta\).
The trap is \(\sec\theta\), which would be right in the first or fourth quadrant but is negative here. 1 + tan θ and 2tan θ add the parts instead of combining them as a square root.
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