6. In $\triangle A B C$, $\cos A, \sin A, \tan B$ form a geometric sequence with a common ratio of $\frac{3}{4}$, then $\cot C=$ $\qquad$
$\sin A=\frac{3}{4} \cos A \Rightarrow \sin ^{2} A=\frac{9}{16} \cos ^{2} A$, substitute into $\sin ^{2} A+\cos ^{2} A=1$ to get $\left\{\begin{array}{l}\sin ^{2} A=\frac{9}{25}, \\ \cos ^{2} A=\frac{16}{25}\end{array} \Rightarrow\left\{\begin{array}{l}\sin A=\frac{3}{5}, \\ \cos A=\frac{4}{5}\end{array} \Rightarrow \tan A=\frac{3}{4}\right.\right.$. Also, $\tan B=\frac{3}{4} \sin A=\frac{9}{20}$, thus $\tan C=-\tan (A+B)=-\frac{\frac{3}{4}+\frac{9}{20}}{1-\frac{3}{4} \cdot \frac{9}{20}}=-\frac{96}{53} \Rightarrow \cot C=-\frac{53}{96}$.