$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ and $ab+bc+ca=1$ This was the question :

$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ where $a,b,c$ are positive number and $ab+bc+ca=1$ ,
$5({1-a^2\over 1+a^2}+{1-b^2\over 1+b^2}+{1-c^2\over 1+c^2}) =  ?$

I have spent a lot of time trying rearranging the terms and combine the two conditions to get the required expression but I couldn't get the right form.
I wrote it like :
$5({2\over 1+a^2}+{2\over 1+b^2}+{2\over 1+c^2} -3) $
This seemed simpler than the required expression , I could not come up with anything else useful.
Could someone please help me in solving this question ?
Thanks !

The hint.
Let $a=\tan\frac{\alpha}{2},$ $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset(0^{\circ},180^{\circ}).$
Thus, $ab+ac+bc=1$ gives $\alpha+\beta+\gamma=180^{\circ}$ and the rest gives 
$$\frac{3}{\sin\alpha}=\frac{4}{\sin\beta}=\frac{5}{\sin\gamma},$$ which gives $\gamma=90^{\circ}$ and $c=1.$
Can you end it now?
Since $c=1$, we obtain: $$b\in\{2,\frac{1}{2}\}$$ and $$a\in\{3,\frac{1}{3}\},$$
but since $$ab+ac+bc=1,$$ we obtain:
$$(a,b,c)=\left(\frac{1}{3},\frac{1}{2},1\right).$$
