(d)
Given that, $\frac{1 - \cos{2\theta}}{1 + \cos{2\theta}} = 3$
$\Rightarrow \ \frac{2\sin^{2}\theta}{2\cos^{2}\theta} = 3$
$\Rightarrow \ \tan^{2}{\theta = \left( \ \sqrt{3} \right)^{2}}$
$\Rightarrow \ \tan^{2}\theta = \tan^{2}\frac{\pi}{3}\$
$\Rightarrow \ \theta = n\pi \pm \frac{\pi}{3}$