This question came from my solution to this previously asked question
If $A+B+C= 180^{\circ}$ and $\cos A = \cos B \cos C$, then $\tan B \tan C$ is equal to? (the answer is 2).
Although the solutions posted on that page are much simpler than mine, the answer I got sparked a new question I can't seem to solve.
My solution went as follows:
$$\tan(B)\tan(C)=\frac{\sin(B)\sin(C)}{\cos(B)\cos(C)}=\frac{\frac12(\cos(B-C)-\cos(B+C))}{\frac12(\cos(B-C)+\cos(B+C))}=\frac{\cos(B-C)-\cos(B+C)}{\cos(B-C)+\cos(B+C)}$$
$$\cos(B+C)=\cos(180^\circ-A)=-\cos(A)=-\cos(B)\cos(C)\\\cos(B-C)=\cos(B)\cos(C)+\sin(B)\sin(C)$$
$$\tan(B)\tan(C)=\frac{2\cos(B)\cos(C)+\sin(B)\sin(C)}{\sin(B)\sin(C)}=\frac{2}{\tan(B)\tan(C)}+1$$
$$(\tan(B)\tan(C))^2-\tan(B)\tan(C)-2=0$$
Let $u = \tan(B)\tan(C)$
$$u^2-u-2=0\qquad(u-2)(u+1)=0\qquad u=2, -1\implies\tan(B)\tan(C)=2, -1$$
From here, how do I deduce that $\tan(B)\tan(C)=-1$ is not a solution?
You got $$\cos\beta\cos\gamma=\frac{1}{2}(\cos(\beta+\gamma)+\cos(\beta-\gamma))=$$ $$=\frac{1}{2}(-\cos\beta\cos\gamma+\cos\beta\cos\gamma+\sin\beta\sin\gamma)=\frac{1}{2}\sin\beta\sin\gamma,$$ which gives $$\tan\beta\tan\gamma=2.$$ Id est, it can not be equal to $-1$, otherwise we obtain $\sin\beta\sin\gamma=\cos\beta\cos\gamma=0.$
Just in the given must be added $\cos\beta\cos\gamma\neq0$.