The general solution of $\cot \theta + \tan \theta = 2$ is

Question

$\theta = \frac{\pi}{2} + (-1)^n \frac{\pi}{8}$

$\theta = \frac{\pi}{2} + (-1)^n \frac{\pi}{4}$

$\theta = \frac{\pi}{2} + (-1)^n \frac{\pi}{6}$

$\theta = n\pi + (-1)^n \frac{\pi}{8}$

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