(c)
$\frac{1 + \tan\theta}{1 - \tan\theta} = 3\left( \frac{3\tan{\theta - \tan^{3}\theta}}{1 - 3\tan^{2}\theta} \right)$
On simplification, we get
$3\tan^{4}{\theta - 6\tan^{2}{\theta + 8\tan{\theta - 1 = 0}}}$
$\therefore\$Product of roots $= \tan{\alpha \bullet \tan{\beta \bullet \tan{\gamma \bullet \tan{\delta = - \frac{1}{3}}}}}$