Explanation

(c)
Given, $\frac{\sin{(x + 3\alpha)}}{\sin{(\alpha - x)}} = 3$
Applying componendo and dividendo, we get
$\frac{\sin{(x + 3\alpha) + \sin{(\alpha - x)}}}{\sin{(x + 3\alpha) - \sin{(\alpha - x)}}} = \frac{3 + 1}{3 - 1}$
$\Rightarrow \ \ \frac{2\sin{2\alpha\cos{(\alpha + x)}}}{2\cos{2\alpha\sin{(\alpha + x)}}} = 2$
$\Rightarrow \ \ \frac{\tan{2\alpha}}{\tan{(\alpha + x)}} = 2$
$\Rightarrow \ \ \frac{2\tan\alpha}{1 - \tan^{2}\alpha} \times \frac{(1 - \tan{\alpha\tan x})}{(\tan{\alpha + \tan x})} = 2$
$\Rightarrow \ \tan{\alpha - \tan^{2}{\alpha\tan{x = \tan{\alpha + \tan{x - \tan^{3}{\alpha - \tan^{2}{\alpha\tan x}}}}}}}$
$\Rightarrow \ \tan{x = \tan^{3}\alpha}$