If $\text{cos}2B=\frac{\cos (A+C)}{\cos (A-C)}$  then $\text{cotA},\text{ cotB},\text{ cotC}$  are in

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

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

                $\frac{\cancel{2}{\cos }^{2}B}{\cancel{2}{\sin }^{2}B}=\frac{\cancel{2}\cos A\cos C}{\cancel{2}\sin A\sin C}$

                ${\cot }^{2}B=\cot A\cot C$

                $\cot A,\cot B,\cot C$  are in G.P