Explanation

(d)
Given, $3\tan(\theta - 15) = \tan(\theta + 15)$
$\frac{\tan A}{\tan B} = \frac{3}{1},$ where $A = \theta + 15{^\circ},\ B = \theta - 15{^\circ}$
$\Rightarrow \ \ \frac{\tan{A + \tan B}}{\tan{A - \tan B}} = \frac{3 + 1}{3 - 1}$
(Applying componendo and dividendo)
$\Rightarrow \ \ \frac{\sin{(A + B)}}{\sin{(A - B)}} = 2$
$\Rightarrow \ \sin{2\theta = 2\sin{30{^\circ}}}$
$\Rightarrow \ \sin{2\theta = \frac{2.1}{2} = 1 = \sin\frac{\pi}{2}}$
$\Rightarrow \ \ 2\theta = 2n\pi + \frac{\pi}{2}$
$\Rightarrow \ \ \theta = n\pi + \frac{\pi}{4}$