If $\cos 2B=\frac{\cos (A+C)}{\cos (A-C)}$ , then $tanA,tanB,tanC$  are in 

$\frac{\cos 2B}{1}=\frac{\cos (A+C)}{\cos (A-C)}\Rightarrow \frac{1-\cos 2B}{1+\cos 2B}=\frac{\cos (A-C)-\cos (A+C)}{\cos (A-C)+\cos (A+C)}$

( $\because$ By the method of componendo and dividendo)

$$\begin{gathered} \frac{2{\sin }^{2}B}{2{\cos }^{2}B}=\frac{2\sin A\sin C}{2\cos A\cos C} \\ {\tan }^{2}B=\tan A\tan C \end{gathered}$$