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

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(B) Given: $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$Using the expansion formulas for $\cos(A+C)$ and $\cos(A-C)$:$\cos 2B = \frac{\cos A \cos C -

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$G.P.$

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(B) Given: $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$Using the expansion formulas for $\cos(A+C)$ and $\cos(A-C)$:$\cos 2B = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$Divide the numerator and denominator by $\cos A \cos C$:$\cos 2B = \frac{1 - 	an A 	an C}{1 + 	an A 	an C}$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 applying componendo and dividendo or cross-multiplying:$(1 - 	an^2 B)(1 + 	an A 	an C) = (1 + 	an^2 B)(1 - 	an A 	an C)$$1 + 	an A 	an C - 	an^2 B - 	an^2 B 	an A 	an C = 1 - 	an A 	an C + 	an^2 B - 	an^2 B 	an A 	an C$Simplifying the equation:$2 	an A 	an C = 2 	an^2 B$$	an^2 B = 	an A 	an C$Since the square of the middle term is equal to the product of the first and third terms,$	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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