romsek MHF Helper Nov 2013 6,725 3,029 California Dec 26, 2016 #2 ? let's use $\theta$ for angles and $x$ for lengths $\tan(\theta) = \dfrac{1}{\cot(\theta)} = x$ $\tan^{-1}(x) = \theta = \cot^{-1}\left(\dfrac 1 x\right)$ Reactions: 1 person

? let's use $\theta$ for angles and $x$ for lengths $\tan(\theta) = \dfrac{1}{\cot(\theta)} = x$ $\tan^{-1}(x) = \theta = \cot^{-1}\left(\dfrac 1 x\right)$