## Precalculus (10th Edition)

$-0.12$
Note that $\cot {x}=\frac{1}{\tan{(x)}}$. Hence, $\cot^{-1}{x}=\tan^{-1} {\left(\frac{1}{x}\right)}$. Use a calculator in radian mode and round the result to decimal places to obtain $\cot^{-1}{(-8.1)}=\tan^{-1} \left({\frac{1}{-8.1}}\right)\approx-0.12$