In $\triangle ABC$ with $A = \frac{\pi}{4}$, what is the range of $\tan B\tan C$? 
In a $\triangle ABC\;,$ If $\displaystyle A=\frac{\pi}{4}\;,$ and $\tan B\cdot \tan C = p\;,$ Then range of $p$

$\bf{My\; Try::}$ For a $\triangle ABC\;, A+B+C=\pi.$ So we get $\displaystyle A+B=\frac{3\pi}{4}$
So $$\tan(A+B)=-1\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B} = -1$$
So we get $$\tan A+\tan B=p-1$$
Now how can i solve after that help required, Thanks

Like Triangular sides,   if $\tan B=q,\tan C=\dfrac{q+1}{q-1}$
So, we have $p=\dfrac{q(q+1)}{q-1}$
$$\iff q^2+q(1-p)+p=0$$
As $q$ is real, the discriminant $$(1-p)^2-4p$$ must be $\ge0$
$$\implies p^2-6p+1\ge0$$
Now if $(x-a)(x-b)\ge0$ with $a\le b,$  we need $x\ge b$ or $x\le a$
