Solving a quadratic for $\tan \theta$ - why not two answers? Given $a =\tan10^\circ $, find $\tan5^\circ $ in terms of $a$  
If I let $\tan5^\circ =t$ then I can use half-angle formula to get to $a =\frac{ 2t}{ 1-t^2 } $
and hence $at^2+2t-a=0$ , but solving this would appear to give $$t ={ -1\pm \sqrt{1+a^2}}\over{ a } $$
So I don't understand why there is only a single answer $$t =\frac{ -1+ \sqrt{1+a^2}}{ a } $$
I obviously have to discount the negative root but I can't quite see why - what am I missing?Thank you!

Observe that  $t_1(=\tan5^\circ), t_2$ are the two roots of $$at^2+2t-a=0$$
$$t_1t_2=\dfrac{-a}a=-1\implies t_2=-\dfrac1{t_1}=-\cot5^\circ=\tan(90^\circ+5^\circ)$$
This  is due to the fact if $$\tan2x=\tan2A$$
$$2x=180^\circ n+2A$$ where $n$ is any integer
$x=90^\circ n+A$ where $n\equiv0,1\pmod2$
