Explanation

(c)
Given, $\tan^{- 1}{2\theta + \tan^{- 1}{3\theta = \frac{\pi}{4}}}$
$\therefore\ \ \ \tan^{- 1}{\left( \frac{2\theta + 3\theta}{1 - 2\theta \times 3\theta} \right) = \tan^{- 1}1}$
$\Rightarrow \ \ \ 6\theta^{2} + 5\theta - 1 = 0$
$\Rightarrow \ \ \theta = \frac{- 5 \pm \sqrt{25 + 24}}{2 \times 6}$
$= \frac{- 5 \pm 7}{12} = - 1,\frac{1}{6}$
$\Rightarrow \ \ \theta = \frac{1}{6}$