If A + B + C = pi and cos A = cos B cos C, th | vedclass

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(B) Given $A + B + C = \pi,$ so $A = \pi - (B + C).$Taking cosine on both sides,$\cos A = \cos(\pi - (B + C)).$Since $\cos(\pi - 	heta) = -\cos 	het

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$2$

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(B) Given $A + B + C = \pi,$ so $A = \pi - (B + C).$Taking cosine on both sides,$\cos A = \cos(\pi - (B + C)).$Since $\cos(\pi - 	heta) = -\cos 	heta,$ we have $\cos A = -\cos(B + C).$Given $\cos A = \cos B \cos C,$ we substitute this into the equation:$-\cos(B + C) = \cos B \cos C.$Using the expansion formula $\cos(B + C) = \cos B \cos C - \sin B \sin C,$$-(\cos B \cos C - \sin B \sin C) = \cos B \cos C.$$-\cos B \cos C + \sin B \sin C = \cos B \cos C.$$\sin B \sin C = 2 \cos B \cos C.$Dividing both sides by $\cos B \cos C$ (assuming $\cos B 
eq 0$ and $\cos C 
eq 0$),$\frac{\sin B \sin C}{\cos B \cos C} = 2.$$	an B 	an C = 2.$

If $A + B + C = \pi$ and $\cos A = \cos B \cos C,$ then $\tan B \tan C$ is equal to

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