Explanation

(b)
Given, $\frac{\tan{3A}}{\tan A} = a$
$\Rightarrow \ \frac{3\tan{A - \tan^{3}A}}{\tan{A\left( 1 - 3\tan^{2}A \right)}} = a$
$\Rightarrow \ \ 3 - \tan^{2}{A = a - 3a\tan^{2}A}$
$\Rightarrow \ \tan^{2}{A(3a - 1) = a - 3}$
$\Rightarrow \ \tan{A = \pm \sqrt{\frac{a - 3}{3a - 1}}}$
Now, $\frac{\sin{3A}}{\sin A} = \frac{3\sin{A - 4\sin^{3}A}}{\sin A}$
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$= 3 - 4\sin^{2}{A = 3 - 4\left( \frac{a - 3}{4(a - 1)} \right)}$
$= \frac{3a - 3 - a + 3}{(a - 1)} = \frac{2a}{(a - 1)}$