In $∆ABC$ if $\mathrm{TanB}=\frac{2\mathrm{SinASinC}}{\mathrm{Sin}(\mathrm{A}+\mathrm{C})}$ then $\mathrm{TanA},$$\mathrm{TanB}, \mathrm{TanC}$ are in

$\begin{array}{lll}\tan \mathrm{B}\begin{array}{ll}=& \frac{2\sin \mathrm{C} \sin \mathrm{A}}{\sin (\mathrm{A}+\mathrm{C})}\end{array}& & \\ \frac{\sin (\mathrm{A}+\mathrm{C})}{\sin \mathrm{A} \sin \mathrm{C}}\begin{array}{ll}=& \frac{2}{\tan \mathrm{B}}\end{array}& & \\ \frac{\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C}}{\sin \mathrm{A} \sin \mathrm{C}}& =& \frac{2}{\tan \mathrm{B}}\\ \frac{1}{\tan \mathrm{A}}+\frac{1}{\tan \mathrm{C}}\begin{array}{ll}=& \frac{2}{\tan \mathrm{B}}\end{array}& & \\ \tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} \mathrm{are} \mathrm{in} \mathrm{H}.\mathrm{P}.& & \end{array}$