If $\frac{\tan (\ln 6)\cdot \tan (\ln 2)\cdot \tan (\ln 3)}{\tan (\ln 6)-\tan (\ln 2)-\tan (\ln 3)}=\mathrm{k}$ then the value of k is

$\ln 6=\ln 3+\ln 2$

$\begin{array}{l}\mathrm{if} \mathrm{\gamma }=\mathrm{\alpha }+\mathrm{\beta },\text{ then }\\ \begin{array}{l}\tan \mathrm{\gamma }-\tan \mathrm{\alpha }-\tan \mathrm{\beta }=\tan \mathrm{\gamma }\cdot \tan \mathrm{\alpha }\cdot \tan \mathrm{\beta }\\ \frac{\tan (\ln 6)\cdot \tan (\ln 2)\cdot \tan (\ln 3)}{\tan (\ln 6)-\tan (\ln 2)-\tan (\ln 3)}=1\end{array}\end{array}$