# How do you calculate tan(cot^-1 (9))?

$\tan \left({\cot}^{- 1} \left(9\right)\right) = \frac{1}{9}$
Let $\cot \theta = 9$
Hence $\theta = {\cot}^{- 1} \left(9\right)$
and $\tan \left({\cot}^{- 1} \left(9\right)\right) = \tan \theta = \frac{1}{\cot} \theta = \frac{1}{9}$