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

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(B) Given $A + B + C = \pi$,so $A = \pi - (B + C)$.Substituting this into the given equation $\cos A = \cos B \cos C$:$\cos(\pi - (B + C)) = \cos B \c

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$2$

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(B) Given $A + B + C = \pi$,so $A = \pi - (B + C)$.Substituting this into the given equation $\cos A = \cos B \cos C$:$\cos(\pi - (B + C)) = \cos B \cos C$Using the identity $\cos(\pi - 	heta) = -\cos 	heta$:$-\cos(B + C) = \cos B \cos C$Expanding $\cos(B + C)$:$-(\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 $\cos B, \cos C 
eq 0$):$\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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