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

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(C) Given,$A+B+C=\pi$,which implies $B=\pi-(A+C)$.Given the equation $\cos B=\cos A \cos C$.Substituting $B$,we get $\cos(\pi-(A+C))=\cos A \cos C$.Us

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$2$

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(C) Given,$A+B+C=\pi$,which implies $B=\pi-(A+C)$.Given the equation $\cos B=\cos A \cos C$.Substituting $B$,we get $\cos(\pi-(A+C))=\cos A \cos C$.Using the identity $\cos(\pi-	heta)=-\cos 	heta$,we have $-\cos(A+C)=\cos A \cos C$.Expanding the left side,$-(\cos A \cos C - \sin A \sin C) = \cos A \cos C$.$-\cos A \cos C + \sin A \sin C = \cos A \cos C$.Rearranging the terms,$\sin A \sin C = 2 \cos A \cos C$.Dividing both sides by $\cos A \cos C$,we get $\frac{\sin A \sin C}{\cos A \cos C} = 2$.Therefore,$	an A 	an C = 2$.

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

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