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 $
Best Answer
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.