Explanation

(a)
$\tan{A = \frac{1 - \cos B}{\sin B}}$
$= \frac{2\sin^{2}{(B/2)}}{2\sin{\left( \frac{B}{2} \right)\cos{(B/2)}}}$
$\Rightarrow \ \tan{A = \tan\frac{B}{2}}$
Now, $\tan{2A = \frac{2\tan A}{1 - \tan^{2}A} = \frac{2\tan{(B/2)}}{1 - \tan^{2}{(B/2)}}}$
$= \frac{2\sin{(B/2)\ \cos\ (B/2)}}{\cos^{2}{(B/2) - \sin^{2}{(B/2)}}}$
$= \frac{\sin B}{\cos B}$
$\Rightarrow \ \tan{2A = \tan B}$