Rearrange $y = x\tan y$ to solve for $y$ given $x$

Question

I have an equation: $$x = \frac{y}{\tan y}$$ or rewritten as: $$y = x\tan y$$

How can I rearrange this so that I can calculate $y$ given $x$?

Answer 1

$\def\tanc{\operatorname{tanc}}$ Using Lagrange reversion

$$\tanc(z)=a\iff z=\left(k+\frac12\right)\pi+\frac i2\ln\left(\frac{2 i}{az-i}+1\right)= \left(k+\frac12\right)\pi+\sum_{n=1}^\infty\left(\frac i2\right)^n\left.\frac{d^{n-1}}{dz^{n-1}}\frac{\ln^n\left(\frac{2 i}{az-i}+1\right)}{n!}\right|_{\left(k+\frac12\right)\pi}$$

Applying a Stirling S1 series:

$$\frac{d^{n-1}}{dz^{n-1}}\frac{\ln^n\left(\frac{2 i}{az-i}+1\right)}{n!}=\sum_{m=0}^\infty \frac{(2i)^{n+m}S_{n+m}^{(n)}}{(n+m)!}\frac{d^{n-1}(a z-i)^{-(n+m)}}{dz^{n-1}}\mathop=^{\left|\frac{2i}{az-i}\right|<1}\sum_{m=0}^\infty\frac{S_{n+m}^{(n)}\Gamma(2n+m-1)(-1)^{n+1}a^{n-1}(2 i)^{n+m}}{(m+n)!\Gamma(m+n)(az-i)^{2n+m-1}}$$

and finally use $\sum\limits_{n=1}^\infty \sum \limits_{m=n}^\infty a_{m,n}=\sum_ \limits{m=1}^\infty\sum \limits_{n=0}^ma_{m,n}$ and the Gauss hypergeometric function:

$$\bbox[3px,border: 1px solid blue]{\begin{align}\tanc_k^{-1}(z)\mathop=^{\left|\frac1{\pi i k z-1}+1\right|<\frac12}\pi k+\sum_{m=1}^\infty\sum_{n=0}^m\frac{S_m^{(n)}\Gamma(n+m-1)i^n 2^{m-n} (\pi k)^m}{\left(\frac i z-\pi k\right)^{n+m-1}\Gamma(m)m!}\,_2\operatorname F_1\left(1-n,-m;2-m-n;1-\frac i{\pi k z}\right)\mathop=^{2<\left|\left(k+\frac12\right)\pi z-i\right|}\left(k+\frac12\right)\pi-\sum_{m=1}^\infty\sum_{n=0}^m\frac{S_m^{(n)}\Gamma(n+m-1)(-1)^ni^{m+n}2^{m-n}z^{n-1}}{\left(z\left(k+\frac12\right)\pi-i\right)^{n+m-1} m!\Gamma(m)}\end{align}}$$

Both cases are shown here and, if infinite, sums are interchangeable. The other series was derived similarly from a slightly modified Lagrange reversion setup. The $k=0$ branch may have convergence problems and the sum gradually converges more slowly as $k\to\infty$ for the simpler series.

Answer 2

Doing so would involve calculating the inverse of the tanc function,

$$y = \operatorname{tanc}^{-1}\frac{1}{x}$$

There is no expression for this function in terms of other elementary functions, so you can't find an analytic solution. You could find the solutions of your equation numerically, though, once you're given an explicit numerical value for $x$.