<s>[INST] Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and  $C$  are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives  $ p \le 3 - 2\sqrt2 $. Or  $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B  = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know   $ \cos (B+C)  =  1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$.  $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives 
$ P \lt 0 $  or $ p \ge 3+ 2\sqrt2 $
 [/INST] Let $C\leq B$ and $B+C=\frac{3\pi}{4}$.
i) When $\frac{3\pi}{8}\leq B<\frac{\pi}{2}$, then
$\frac{\pi}{4}<C\leq \frac{3\pi}{8}$. 
Define $$f=\tan\ B\tan\ ( \frac{3\pi}{4}-B)$$ The range of $f$ is
$\{ \tan^2\frac{3\pi}{8} \leq t<\infty\}$, by (1) continuity, (2)
considering $B\rightarrow \frac{\pi}{2},\ B\rightarrow
\frac{3\pi}{8}$ and (3)
\begin{align*} f' &=\frac{-\cos\ (2B-\frac{\pi}{4} ) }{\sqrt{2}\cos^2 B\sin^2
 (B-\frac{\pi}{4}) } >0
\end{align*} when $\frac{3\pi}{8}<B$
ii) When $\frac{\pi}{2} < B<\frac{3\pi}{4}$, then $
-\infty<\tan\ B<T<0$ for some $T$. And $0<\tan\ C<1$ so that range
of $f$ contains $\{-\infty < t<0\}$.
Hence note that these $B,\ C$ can be angles in a triangle so that $
\{ \tan^2\frac{3\pi}{8} \leq t<\infty\ {\rm or}\ -\infty<t<0\}$.
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