If cos 2B = frac{cos(A + C)}{cos(A - C)},then | vedclass

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(B) Given $\cos 2B = \frac{\cos(A + C)}{\cos(A - C)}$.Using the expansion of $\cos(A+C)$ and $\cos(A-C)$,we get $\cos 2B = \frac{\cos A \cos C - \sin

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$G.P.$

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(B) Given $\cos 2B = \frac{\cos(A + C)}{\cos(A - C)}$.Using the expansion of $\cos(A+C)$ and $\cos(A-C)$,we get $\cos 2B = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$.Dividing the numerator and denominator by $\cos A \cos C$,we get $\cos 2B = \frac{1 - 	an A 	an C}{1 + 	an A 	an C}$.We know that $\cos 2B = \frac{1 - 	an^2 B}{1 + 	an^2 B}$.Equating the two expressions: $\frac{1 - 	an^2 B}{1 + 	an^2 B} = \frac{1 - 	an A 	an C}{1 + 	an A 	an C}$.By componendo and dividendo or cross-multiplication,we get $1 - 	an^2 B + 	an A 	an C - 	an^2 B 	an A 	an C = 1 + 	an^2 B - 	an A 	an C - 	an^2 B 	an A 	an C$.This simplifies to $2 	an^2 B = 2 	an A 	an C$,which implies $	an^2 B = 	an A 	an C$.Thus,$	an A, 	an B, 	an C$ are in $G.P.$

If $\cos 2B = \frac{\cos(A + C)}{\cos(A - C)}$,then $\tan A, \tan B, \tan C$ are in

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