We know that transcendent equations $$\tan{x}=bx$$ and $$x\tan{x}=b$$ can not be solved exactly. But what I concerned most is the relationship between their non-trival roots $x_{n}^{(1)}$ and $x_{n}^{(2)}$, where $x_{n}^{(1)}$ and $x_{n}^{(2)}$ refer to the $n$-th positive solution to the first and second of these equations respectively. For example $x_{n}^{(1)}=x_{n}^{(2)}+\pi/2$( which is incorrect). Any suggestions?
Similar question is the roots of $W(x)\exp(W(x))=x$ and $M(x)\ln(M(x))=x$.
On
When the roots are close to $P_n=(n+\frac12)\pi$, we can approximate the tangent as $$\tan(P_n+h)\approx-\frac1h,$$ and the equations become $$-\frac1h=b(P_n+h),$$ $$-\frac1{h'}(P_{n'}+h')=b.$$ Eliminating $b$, this gives the equation of a cubic $$-\frac1h=-\frac1{h'}(P_{n'}+h')(P_n+h),$$ $$h'=h(P_{n'}+h')(P_n+h).$$ For large $n$ and $n'$, it simplifies to $$h'=hP_{n'}P_n,$$ so that $$\frac{x_n^{(2)}-P_n}{x_n^{(1)}-P_n}\approx P_n^2.$$ Nothing really exciting as this is just an approximation. But my guess is that if a simple relation between the roots had existed, the linearized analysis would have revealed it.
On
In this article you can find the exact solutions of both your equations using an extension of Riemann method http://www4.ncsu.edu/~ces/pdfversions/52.pdf
In the particular case $b=1$, it is known that the roots of $\tan(x_n)=x_n$ are close to : $x_n=(n-\frac{1}{2})\pi-\frac{1}{(n-\frac{1}{2})\pi}-\frac{2}{3}(\frac{1}{(n-\frac{1}{2})\pi})^3-\frac{13}{15}(\frac{1}{(n-\frac{1}{2})\pi})^5-\frac{146}{105}(\frac{1}{(n-\frac{1}{2})\pi})^7-\frac{781}{315}(\frac{1}{(n-\frac{1}{2})\pi})^9-...$
The trivial root $x=0$ is not included in the list. The first positive root $x_1$ is not accurate. The accuracy increasses quickly for larger $\vert n \vert$. The first negative root is denoted $x_0$ ( not accurate) and the next negative roots correspond to $n<0$ so that $x_{-n}=-x_{n+1}$
With the same method of series expansion, I got the roots of $\tan(x_n)=b x_n$ :
$x_n=(n-\frac{1}{2})\pi-\frac{1}{b}\frac{1}{(n-\frac{1}{2})\pi}-\frac{3b-1}{3b^3}(\frac{1}{(n-\frac{1}{2})\pi})^3-\frac{30b^2-20b+3}{15b^5}(\frac{1}{(n-\frac{1}{2})\pi})^5$ $-\frac{525b^3-252b^2+161b-15}{105b^7}(\frac{1}{(n-\frac{1}{2})\pi})^7-\frac{4410b^4-5880b^3+2744b^2-528b+35}{315b^9}(\frac{1}{(n-\frac{1}{2})\pi})^9-...$
and the roots of $x_n \tan(x_n)=b$ with $n>1$ :
$x_n=(n-1)\pi+\frac{b}{(n-1)\pi}-\frac{b^2(b+3)}{3}(\frac{1}{(n-1)\pi})^3+\frac{b^3(3b^2+20b+30)}{15}(\frac{1}{(n-1)\pi})^5$ $-\frac{b^4(15b^3+161b^2+525b+525)}{105}(\frac{1}{(n-1)\pi})^7+\frac{b^5(35b^4+528b^3+2744b^2+5880b+4410)}{315}(\frac{1}{(n-1)\pi})^9-...$
Case $n=1$ : The serie is not convergent for the first root $(n=1)$ because $x_1$ is equivalent to $\sqrt{b}$ when $x$ tends to $0$. So, the serie cannot be expressed in terms of integer powers of $b$ as it was above. An approximate for the first root is :
$x_1=\sqrt b \big( 1-\frac{1}{6}b+ \frac{11}{360}b^2- \frac{17}{5040}b^3- \frac{281}{604800}b^4+ \frac{44029}{119750400}b^5- \frac{12147139}{130767436800}b^6$ $+ \frac{2030761}{784604620800}b^7+ \frac{1381516351}{313841848320000}b^8+…\big)$
This serie is slowly convergent. So, a large number of terms are required if $b$ is not small.