# If $A+B+C= 180^{\circ}$ and $\cos A = \cos B \cos C$, then $\tan B \tan C$ is equal to?

If $A+B+C= 180^{\circ}$ and $\cos A = \cos B \cos C$, then $\tan B \tan C$ is equal to?

(a) $1/2$

(b) $2$

(c) $1$

(d) $-1/2$

I can't figure out the solution, maybe I am missing an onward trick which I am unable to spot. I tried using the fact that $tan A = - tan(B+C)$, used the $tan(x+y)$ formula too, but unfortunately, it does not lead me anywhere. Any help would be appreciated!

• A rather different, but no less interesting question is: how many triples (A, B, C) are there satisfying these relations? – Vincent Jul 7 '18 at 9:37

$$\cos (180-B-C) = \cos B \cos C\implies -\cos (B+C) =\cos B\cos C$$

Now we have $$-\cos B\cos C +\sin B \sin C = \cos B\cos C$$

Can you finish?

• How did I miss that? :( – MathDude3013 Jul 7 '18 at 6:43

Alternatively (your attempt): \begin{align} \tan(B+C)&=\frac{\tan B+\tan C}{1-\tan B\tan C} \Rightarrow \\ 1-\tan B\tan C&=\frac{\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}}{\tan(180^\circ-A)} =\\ &=\frac{\sin (B+C)}{\cos B\cos C\cdot (-\tan A)} = \\ &=\frac{\sin A}{\cos A\cdot (-\tan A)} = \\ &=-1 \Rightarrow \\ \tan B\tan C&=2.\end{align}

• Aah, I was unable to see this. Thanks! – MathDude3013 Jul 7 '18 at 10:08

All possibilities are wrong!

Try $$A=270^{\circ}$$, $$B=-90^{\circ}$$ and $$C=0^{\circ}.$$