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

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$A + B + C = \pi$
Also $A = \pi/4\Rightarrow B + C = 3 \pi/4\Rightarrow 0 < B, C < 3\pi/4$
Now $\tan B \tan C = P\Rightarrow \dfrac{\sin B \sin C}{\cos B \cos C} = p/1$
Applying componendo and dividendo, we get
$$\dfrac{\sin B \sin C + \cos B \cos C}{\cos B \cos C – \sin B \sin C} = \dfrac{1+p}{1-p}$$
$$\Rightarrow \dfrac{\cos (B – C)}{\cos (B + C)} = \dfrac{1 + p}{1 – p}$$
$$\Rightarrow \cos (B – C) = \frac{1 + p}{1 – p} (-1/\sqrt{2}) . . . . . . . . . . . . . (1) [\because B + C = 3 \pi/4]$$
Now, as $B$ and $C$ can vary from $0$ to $3\pi/4$
$$\therefore 0\leq B – C < 3\pi/4\Rightarrow 1/\sqrt{2} < \cos (B – C)\leq 1$$
From $eq” (1)$ substituting the value of $\cos (B – C)$, we get
$$-1/\sqrt{2} < 1 + p/\sqrt{2}(p – 1) \geq 1 \Rightarrow -1/\sqrt{2} < 1 + p/\sqrt{2}(p – 1) and 1 + p/\sqrt{2}(p – 1)\leq 1$$
$$\Rightarrow 0 < 1 + \dfrac{p + 1}{p – 1} and (p + 1) - \dfrac{\sqrt{2} (p – 1)}{\sqrt{2} (p – 1)}\leq 0$$
$$\Rightarrow \dfrac{2p}{p – 1} > 0 and p + 1 - \dfrac{\sqrt{2} p + \sqrt{2}}{\sqrt{2}(p – 1)}\leq 0$$
$$\Rightarrow p (p – 1) > 0 and \dfrac{(1 - \sqrt{2}) p + (\sqrt{2} + 1)}{(p – 1)}\leq 0$$
$$\Rightarrow p\in ( -\infty, 0)\cup (1, \infty), and –p + (\sqrt{2} + 1)2/(p – 1)\leq 0$$
$$\Rightarrow [p – (3 + 2\sqrt{2})] [p – 1]\geq 0$$
Combining the two cases, we get
$$p\in (-\infty, 0)\cup [3 + 2\sqrt{2}, \infty)$$
