Simple trigonometry problem It is given that, $ A+B +C=  \pi $,and $\cos A = \cos B \times \cos C$
I have to prove:  $\tan B \times  \tan C= 2$ 
to prove that, this is what I did:
$$\frac{\sin B}{\cos B} \times \frac{\sin C}{\cos C}=\frac{\sin B\times \sin C }{\cos A} =??$$ 
what should I do now to get the result 2?
 A: $tanBtanC=\frac{sinBsinC}{cosBcosC}=\frac{sinBsinC}{cosA}$
$=\frac{sinBsinC}{cos(\pi-(B+C))}=\frac{sinBsinC}{-cos(B+C)}=\frac{-sinBsinC}{cosBcosC-sinBsinC}=\frac{-\frac{sinBsinC}{cosBcosC}}{1-\frac{sinBsinC}{cosBcosC}}$
$\Rightarrow tanBtanC=\frac{-tanBtanC}{1-tanBtanC}$
Put $tanBtanC=t$
$t=\frac{-t}{1-t}$
solving we get,$t=0$ or $t=2$
A: HINT: we have
$$\frac{cos(\pi-B-C)}{\cos(B)\cos(C)}=\frac{-\cos(B)\cos(C)+\sin(B)\sin(C)}{\cos(B)\cos(C)}=-1+\tan(B)\tan(C)=1$$
A: put $A=\pi - B - C$ in $ cosA =cosB * cosC$ will get $-cos(B+C)=cos(B)cos(C)$ followed by $-1+tan(B)tan(C)=1$ hence giving $tan(B)tan(C)=2$ 
A: Since $A=\pi-(B+C)$ we have:
$$
\cos B \cos C= \cos A=- \cos(B+C)=\sin B \sin C -\cos B \cos C \Rightarrow
$$
$$
\sin B \sin C= 2 \cos B \cos C
$$
Now substitute in the tangents $\tan \alpha= \dfrac{\sin \alpha}{\cos \alpha}$ 
A: Given that $$\cos A=\cos B\cos C$$ Now, substituting $A=\pi-(B+C)$ we get $$\cos (\pi-(B+C))=\cos B\cos C$$ $$\implies -\cos (B+C)=\cos B\cos C$$ $$\implies -(\cos B\cos C-\sin B\sin C)=\cos B\cos C$$ $$\implies \sin B\sin C=2\cos B\cos C$$ $$\implies \frac{\sin B\sin C}{\cos B\cos C}=2$$$$\implies \frac{\sin B}{\cos B}\times \frac{\sin C}{\cos C}=2$$ $$\implies \color{blue}{\tan B\times \tan C=2}$$
