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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!

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    $\begingroup$ A rather different, but no less interesting question is: how many triples (A, B, C) are there satisfying these relations? $\endgroup$ – Vincent Jul 7 '18 at 9:37
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$$\cos (180-B-C) = \cos B \cos C\implies -\cos (B+C) =\cos B\cos C$$

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

Can you finish?

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  • $\begingroup$ How did I miss that? :( $\endgroup$ – MathDude3013 Jul 7 '18 at 6:43
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Alternatively (your attempt): $$\begin{align} \tan(B+C)&=\frac{\tan B+\tan C}{1-\tan B\tan C} \Rightarrow \\ 1-\tan B\tan C&=\frac{\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}}{\tan(180^\circ-A)} =\\ &=\frac{\sin (B+C)}{\cos B\cos C\cdot (-\tan A)} = \\ &=\frac{\sin A}{\cos A\cdot (-\tan A)} = \\ &=-1 \Rightarrow \\ \tan B\tan C&=2.\end{align}$$

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  • $\begingroup$ Aah, I was unable to see this. Thanks! $\endgroup$ – MathDude3013 Jul 7 '18 at 10:08
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All possibilities are wrong!

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

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