Explanation

(b)
Given that, $\cos{2B = \frac{\cos(A + C)}{\cos(A - C)}}$
$= \frac{\cos{A\cos{C - \sin{A\sin C}}}}{\cos{A\cos{C + \sin{A\sin C}}}}$
$\Rightarrow \frac{1 - \tan^{2}B}{1 + \tan^{2}B} = \frac{1 - \tan{A\tan C}}{1 + \tan{A\tan C}}$
$\Rightarrow 1 + \tan^{2}{B - \tan{A\tan{C - \tan{A\tan{C\tan^{2}B}}}}}$
$= 1 - \tan^{2}{B + \tan{A\tan{C - \tan{A\tan{C\tan^{2}B}}}}}$
$\Rightarrow 2\tan^{2}{B = 2\tan{A\tan C}}$
$\Rightarrow \tan^{2}{B = \tan{A\tan C}}$
Hence, $\tan{A,\tan{B\ }}$and $\tan{C\ }$will be in GP