Approximating the first +ve root of $\tan(\lambda)= \frac{a\lambda+b}{\lambda^2-ab}$, $\lambda\in(0,\pi/2)$ I have this expression, which I wish to approximate the solution of:
$$\tan(\lambda)= \frac{a\lambda+b}{\lambda^2-ab}$$
where $a,b>0$. It can be assumed that $ab<\pi/2$, therefore there will always exist a +ve solution for lambda in $(0,\pi/2)$
Is there a way to approximate $\tan(\lambda)$ for $\lambda\in(0,\pi/2)$ aside from the following expansion?
$$\tan(\lambda) = \lambda + \lambda^3/3 + 2\lambda^5/15 + O(\lambda^7)$$
This approximation breaks down pretty quickly in the interval $\lambda\in(0,\pi/2)$, additionally, the analytic solutions for it aren't really very tractable, is there another way I can approximate this solution in terms of $a$ and $b$? Perhaps another approximation for $\tan$?

EDIT I: 
Ok, so
$$\tan(\lambda)\simeq \frac{\lambda}{1-\lambda^2/3}$$
seems like a superior approximation (see plot below), however again pretty nightmarish.

EDIT II:
Using the above approximation one can generate the following
$$0 = (a+3)\lambda^3 + b\lambda^2 - 3a(1+b)\lambda - 3b$$
This may be able to be approximated, $\lambda\in(\sqrt{ab},\pi/2)$, and I believe typically $a\sim 10^{-1}$ and $b\sim 10^0$.
 A: It's easy enough to show $\sqrt{a b} < \lambda$, and the intersection of the two curves is transverse (i.e., the difference, $\tan \lambda - \frac{a \lambda + b}{\lambda^2 - a b}$, is monotonically increasing on $\left( \sqrt{a b}, \frac{\pi}{2} \right)$), so try Newton's method, starting from the midpoint of the interval.  \begin{align*}
x_0 &= \frac{1}{2} \left( \sqrt{a b} + \frac{\pi}{2} \right)  \\
x_{i+1} &= x_i - \frac{(a b - x_i^2)(b + a x_i + (a b - x_i^2))\tan(x_i)}{a^2 b + 2 b x_i + a x_i^2 + (a b - x_i^2)^2 \sec^2(x_i)}
\end{align*}
I'm short of time, but I'd bet a penny that $x_3$ will beat your 90-95% accuracy requirement for all valid choices of $a$ and $b$.

Not so good for $a \sim 100, b \sim 10^{-3}$ (because for these parameters, Newton's method tends to jump to a solution further to the right, which I suppose I could fix with some fencing).  However, for $a \sim 10^{-1}, b \sim 1$, some Monte Carlo sampling suggests $x_2$ and $x_3$ agree with $\lambda$ within about 3%.  ($x_1$ is only within about 10%.)  Worst agreement occurs for $a \ll b$ or $a \gg b$ and could probably be fixed by also running a Newton's method from $x_0 = 1.5$, since the intersection is pressed up against the side of the asymptote.
