समाधान y=tan(1divx)

$$y = \tan ( 1 \div x )$$

x को लागि हल गर्नुहोस्

$$x=\left(\pi +2n_{4}\pi +arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\right)^{-1}\text{, }n_{4}\in \mathrm{Z}\text{, }\exists n_{43}\in \mathrm{Z}\text{ : }\left(n_{4}>-\frac{1}{4}+\left(-\frac{1}{2}\right)\pi ^{-1}arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+\frac{1}{2}n_{43}\text{ and }n_{4}<\frac{1}{4}+\left(-\frac{1}{2}\right)\pi ^{-1}arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+\frac{1}{2}n_{43}\right)$$

$$x=\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{23}\right)^{-1}\text{, }n_{23}\in \mathrm{Z}\text{, }\left(\left(\left(not(n_{23}=\left(-\frac{1}{2}\right)arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\pi ^{-1})\text{ and }\exists n_{43}\in \mathrm{Z}\text{ : }\left(n_{43}<\pi ^{-1}arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})-\frac{1}{2}+2n_{23}\text{ and }n_{43}>\pi ^{-1}arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})-\frac{3}{2}+2n_{23}\right)\right)\text{ and }\left(not(n_{23}=\left(-\frac{1}{2}\right)arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})\pi ^{-1})\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{23}\right)^{-1}=2\pi ^{-1}\left(1+2n_{1}\right)^{-1}\right)\right)\text{ and }\left(not(\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{23}\right)^{-1}=0)\text{ or }arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{23}=0\right)\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\left(arcSin(y\left(y^{2}+1\right)^{-\frac{1}{2}})+2\pi n_{23}\right)^{-1}=2\pi ^{-1}\left(1+2n_{1}\right)^{-1}$$

y को लागि हल गर्नुहोस्

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

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