Show $\frac{\sin20^\circ\sin30^\circ\tan40^\circ}{\sin20^\circ\sin30^\circ+\tan40^\circ\sin50^\circ} =\tan10^\circ$ The question arose in finding $\alpha$ in this diagram:

(source: mei.org.uk)
With a bit of work with the sine rule it can be shown that:
$$\tan(\alpha) = \frac{\sin(20^\circ)\sin(30^\circ)\tan(40^\circ)}{\sin(20^\circ)\sin(30^\circ)+\tan(40^\circ)\sin(50^\circ)}$$
... and this can be evaluated on a calculator to find $\alpha = 10^\circ$. But I'm not satisfied in using a calculator.

I want to know if there's a neat way of manipulating the expression on the RHS so that it simplifies out to $\tan(10^\circ)$.

I've had a good go at it - using double/triple/quadruple angle formulae to re-write everything in terms of $\sin(10^\circ)$, $\cos(10^\circ)$ and $\tan(10^\circ)$ and trying to simplify - but just seem to make things more complicated.
Can you find a nice way of doing it?
 A: Divide the numerator and denominator by $\sin20\sin30= \frac12\sin20$
\begin{align}
&\frac{\sin20\sin30\tan40}{\sin20\sin30+\tan40\sin50}\\
=& \frac{\sin40}{\cos40+4\sin50\cos20}
= \frac{\sin40}{(\cos40+1)+2\cos20}
=\frac{\sin20}{\cos20+1}= \tan10\\
\end{align}
where the identities $\sin2t=2\sin t\cos t$, $\cos2t+1=2\cos^2t$ are used twice each.
A: Just for the fun of it, consider the equation
$$\frac{\sin (2 x) \sin (3 x) \tan (4 x)}{\sin (2 x) \sin (3 x)+\sin (5 x) \tan (4
   x)}=\tan(x)$$
Let $x=\tan^{-1}(t)$ and this write
$$\frac{4 (t-1) (t+1) \left(t^2-3\right) t}{\left(3 t^2-1\right) \left(t^4-10
   t^2+13\right)}+t=0$$ Discarding the trivial $t=0$, this reduces to
$$\frac{3 t^6-27 t^4+33 t^2-1}{\left(3 t^2-1\right) \left(t^4-10 t^2+13\right)}=0$$
The numerator is a cubic in $t^2$ with three real roots since $\Delta=331776$. Using the trigonometric method (!!), they are
$$t^2_k=3+\frac{8}{\sqrt{3}} \cos \left(\frac{\pi }{18}-\frac{2 \pi  k}{3}\right) \qquad \text{for} \qquad k=0,1,2$$ They are all positive and the smallest (corresponding to $k=2$) is
$$t^2_2=3-\frac{8}{\sqrt{3}}\sin \left(\frac{2 \pi }{9}\right)\implies
x=\tan^{-1}\Bigg[\sqrt{3-\frac{8}{\sqrt{3}}\sin \left(\frac{2 \pi }{9}\right) } \Bigg]$$
At this point, I am totally stuck but numerically this is $\frac \pi {18}$
