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

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 $$p \ge 3+ 2\sqrt2$$

1) The second method is wrong because of a silly mistake.

2) The first method is wrong because apart from discriminant, it is also important to note that there are restrictions on the values of $$B$$ and $$C$$ , and thus their are restrictions on $$tan C$$, and $$tan B$$, and thus their are restrictions on $$p$$.

When both $$B$$ and $$C$$ are acute angles, both the roots of the above equations are positive. thus $$p \gt 1$$.

When one of them is obtuse, $$\tan B \tan C \lt 0 .$$

thus $$p \lt 0 .$$

This with the intersection of non-negative discriminant, gives the correct answer.

Which give the range obtained in the third answer.

I think that every approach should give correct solution. Different solutions appear if to do mistakes.

Conditions to $$\angle B$$ and $$\angle C$$ are WLOG $$\begin{cases} \angle B+\angle C = \dfrac{3\pi}4\\[4pt] \tan\angle B\tan\angle C = p\\[4pt] \angle B\in\left(0,\dfrac{3\pi}8\right]\\ \angle C\in\left[\dfrac{3\pi}8,\dfrac{3\pi}4\right).\tag1 \end{cases}$$

Also, are known relations

$$\tan(a+b) = \dfrac{\tan a + \tan b}{1-\tan a \tan b},\tag{2a}$$ $$\tan\dfrac34\pi = -1,\tag{2b}$$ $$\tan\dfrac38\pi = \frac{1+\cos\frac34\pi}{\sin\frac34\pi}=\sqrt2+1.\tag{2c}$$

Let $$x=\tan \angle B$$ and $$y=\tan \angle C.$$ then

$$y = \tan\left(\dfrac{3\pi}4-x\right) = \dfrac{-1-x}{1-x} = \dfrac{x+1}{x-1},\tag3$$ $$xy=p.$$

Therefore, $$P(p,x) = x^2+x-p(x-1) = 0.\tag4$$

If $$\,\underline{x=y=\sqrt2+1}\,$$ then $$P(p,\sqrt2+1) = 4+3\sqrt2 -p\sqrt2 = 0,\\$$ $$p=2\sqrt2+3\tag5.$$

If $$\,\underline{x\in (0, \sqrt2+1)}\,$$ then

$$P(p,z)\,$$ has not roots in the interval $$\,z\in(-1,0)\,$$ and has one root in the interval $$\,z\in(0,\sqrt2+1),\,$$ $$\begin{cases} P(p,-1)P(p,0) > 0\\[4pt] P(p,0)P(p,\sqrt2+1)<0, \end{cases}$$

$$\begin{cases} (p+1)p > 0\\[4pt] p(2\sqrt2+3-p)<0, \end{cases}$$ $$p\in(-\infty,0)\cup(2\sqrt2+3,\infty).$$

Taking in account $$(5),$$ the answer is

$$\color{green}{\boxed{{\phantom{\Big|}\mathbf{p\in(-\infty,0)\cup[2\sqrt2+3,\infty)}.}}\tag6}$$

If the condition $$(6)$$ is satisfied, then the common solution is $$(\angle B,\angle C)\in\{(f_\angle(z_1),f_\angle(z_2)), (f_\angle(z_2),f_\angle(z_1))\},\tag7$$ where $$D=p^2-6p+1 = (p-3-2\sqrt2)(p-3+2\sqrt2),\tag8$$ $$z_{1,2} = \dfrac{p-1\pm\sqrt D}2,\tag9$$ $$f_\angle(z) = \arctan z +\dfrac\pi2(1-\operatorname{sgn} z).\tag{10}$$

Example 1. $$\quad p=-\frac32,\quad z\in\{-3,\frac12\},\quad \angle B\approx 27^\circ,\quad \angle C \approx 109^\circ,\quad\angle B+\angle C = 135^\circ =\frac34\pi.$$

Example 2. $$\quad p=6,\quad z\in\{2,3\},\quad \angle B\approx63^\circ, \angle C \approx 72^\circ,\quad \angle B +\angle C = 135^\circ -\frac34\pi.$$

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}.

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}

ii) When $$\frac{\pi}{2} < B<\frac{3\pi}{4}$$, then $$-\infty<\tan\ B for some $$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.

For $$A,B,C$$ to be the angles of a triangle, not only it shall be $$A+B+C = \pi$$, but also $$0 \le A,B,C$$, or strictly greater than $$0$$ if you exclude the degenerate case.

So, the correct formulation of the problem is $$\left\{ \matrix{ A = \pi /4 \hfill \cr \tan B\tan C = p \hfill \cr A + B + C = \pi \hfill \cr 0 \le A,B,C \hfill \cr} \right.\quad \Rightarrow \quad \left\{ \matrix{ \tan B\tan C = p \hfill \cr B + C = 3\pi /4 \hfill \cr 0 \le B,C\left( { \le 3\pi /4} \right) \hfill \cr} \right.$$

Taking care of this further restiction, whatever approach you follow (correctly) you will get to the unique right solution.

For instance, given $$A=\pi /4$$, we may start from the symmetrical case , isosceles triangle $$B=C=3 \pi /8$$ and put \eqalign{ & \left\{ \matrix{ B = 3\pi /8 + D \hfill \cr C = 3\pi /8 - D \hfill \cr - 3\pi /8 \le D \le 3\pi /8 \hfill \cr \tan \left( {3\pi /8 + D} \right)\tan \left( {3\pi /8 - D} \right) = p \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad \left\{ \matrix{ B = 3\pi /8 + D \hfill \cr C = 3\pi /8 - D \hfill \cr - 3\pi /8 \le D \le 3\pi /8 \hfill \cr t = \tan \left( {3\pi /8} \right) = \sqrt 2 + 1 \hfill \cr - t \le x = \tan D \le t \hfill \cr {{t^2 - x^2 } \over {1 - t^2 x^2 }} = p \hfill \cr} \right. \cr & \Rightarrow \quad \left\{ \matrix{ t^2 = 3 + 2\sqrt 2 \hfill \cr 0 \le x^2 \le t^2 \hfill \cr p(x) = {{t^2 - x^2 } \over {1 - t^2 x^2 }} \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad p \in \left[ {t^2 = 3 + 2\sqrt 2 , + \infty } \right) \cup \left( { - \infty ,0} \right] \cr}

The reason why the 1st method is wrong :

We can rewrite the following equation

$$f(x) = x^2 - (p-1)x + p$$

As we know the sum and product of $$\tan A$$ and $$\tan B$$

Settings discriminant greater than equal to zero.

It seems that you meant "the sum and product of $$\color{red}{\tan C}$$ and $$\tan B$$".

Considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $$0\lt B\lt \frac 34\pi$$.

This means that we have to consider the condition that $$f(x)=0$$ has at least one solution such that $$x\lt -1$$ or $$x\gt 0$$.

Therefore, the 1st method is wrong.

The reason why the 2nd method is wrong :

In the solution in Quora,

tan B + tan C = tanB tanC - 1

tanB+tan(3pi/4 - B) = p

This step is wrong. This should be $$\tan B+\tan\left(\frac 34\pi-B\right)=\color{red}{p-1}$$ Then, we get $$\tan^2B-(p-1)\tan B+p=0$$ But, again, similarly as the 1st method, considering the condition that the discriminant is greater than or equal to zero is not enough because we also have to have $$0\lt B\lt\frac 34\pi$$.

Therefore, the 2nd method is wrong.

To make the two methods correct, we have to consider the condition that $$x^2-(p-1)x+p=0$$ has at least one solution such that $$x\lt -1$$ or $$x\gt 0$$.

Solving this gives $$p\in (-\infty,0)\cup [3+2\sqrt 2,\infty)$$ which is the same as the answer in the third method.

Let $$\tan\beta=t$$.

Thus, $$t\neq1$$, otherwise $$\gamma=\frac{\pi}{2},$$ which is impossible.

Also, $$0<\beta<\frac{3\pi}{4},$$ which is $$0<\beta<\frac{\pi}{4}$$ or $$\frac{\pi}{4}<\beta<\frac{\pi}{2}$$ or $$\frac{\pi}{2}<\beta<\frac{3\pi}{4},$$ which is $$0 or $$t>1$$ or $$t<-1.$$

Consider three cases.

1. $$t>1$$.

By AM-GM we obtain: $$p=\tan\beta\tan\left(\frac{3\pi}{4}-\beta\right)=t\cdot\frac{-1-t}{1-t}=\frac{t^2+t}{t-1}=\frac{t^2+t-2+2}{t-1}=$$ $$=t+2+\frac{2}{t-1}=3+t-1+\frac{2}{t-1}\geq3+2\sqrt{(t-1)\cdot\frac{2}{t-1}}=3+2\sqrt2.$$ The equality occurs for $$t-1=\frac{2}{t-1}$$ and since $$\lim\limits_{t\rightarrow1^+}p=+\infty$$ and $$p$$ is a continuous function on $$(1,+\infty)$$, we got a range of $$p$$ in this case: $$[3+2\sqrt2,+\infty).$$

1. $$0 We obtain: $$p=-\frac{t(1+t)}{1-t}<0.$$ Also, $$\lim\limits_{t\rightarrow1^-}p=-\infty,$$ $$\lim\limits_{t\rightarrow0^+}p=0$$ and $$p$$ is a continuous function on $$(0,1),$$ which says that the range of $$p$$ in this case it's $$(-\infty,0).$$

2. $$t<-1$$.

In this case $$p<0$$, $$\lim_{t\rightarrow-1^-}p=0,$$ $$\lim_{t\rightarrow-\infty}p=-\infty$$ and $$p$$ is a continuous function on $$(-\infty,-1).$$

Thus, the range of $$p$$ in this case it's $$(-\infty,0),$$ which gives the answer: $$(-\infty,0)\cup[3+2\sqrt2,+\infty).$$

We should consider discriminants but in the way they arise directly:

Let $$tx = \tan(x)$$ for shorthand notation

$$t_A= t{(\pi-B-C)}$$

$$1=\frac{-(tA+tB)}{1-tA\, tB}$$

$$tA \,tB=p\quad$$ plug in and simplify to find quadratic equation roots $$tB^2+(1-p)tB+p=0$$ $$2\,tA=-(1-p)-\sqrt{1-6p+p^2}$$ $$2\,tB=-(1-p)+\sqrt{1-6p+p^2}$$

For this to be real, quantity under radical sign should be positive.

The roots of quadratic equation then supply upper and lower limits automatically.

$$p_{lower}= 3+2\sqrt{2}\quad p_{lower}= 3-2\sqrt{2}$$

when $$p$$ should not lie between these limits as the quantity under radical sign is:

$$\sqrt{p^2 -6p +1 } = \sqrt{(p-p_{lower})(p-p_{upper})}$$