Réitigh y=tan1/x

$$y = \tan \frac { 1 } { x }$$

Réitigh do x.

$$x=\frac{1}{2\pi n_{1}+\arcsin(\frac{y}{\sqrt{y^{2}+1}})+\pi }\text{, }n_{1}\in \mathrm{Z}\text{, }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{1}>\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }-1}{4}\text{ and }n_{1}<\frac{2n_{3}-\frac{2\arcsin(\frac{y}{\sqrt{y^{2}+1}})}{\pi }+1}{4}\right)$$

$$x=\frac{1}{2\pi n_{2}+\arcsin(\frac{y}{\sqrt{y^{2}+1}})}\text{, }n_{2}\in \mathrm{Z}\text{, }\left(n_{2}\neq 0\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}\text{bmod}2=1\text{ and }n_{2}\geq \frac{n_{3}}{2}\text{ and }n_{2}\leq \frac{n_{3}}{2}+1\right)\right)\text{ or }\left(y\neq 0\text{ and }\exists n_{3}\in \mathrm{Z}\text{ : }\left(n_{3}\text{bmod}2=1\text{ and }n_{2}\geq \frac{n_{3}}{2}\text{ and }n_{2}\leq \frac{n_{3}}{2}+1\right)\right)$$

Réitigh do y.

$$y=\tan(\frac{1}{x})$$

$$x\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{2}{2\pi n_{1}+\pi }$$