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

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(A) Given $\cos 2B = \frac{\cos(A+C)}{\cos(A-C)}$.Using the formula $\cos 2B = \frac{1-	an^2 B}{1+	an^2 B}$ and expanding the right side:$\frac{1-

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Geometric Progression.

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(A) Given $\cos 2B = \frac{\cos(A+C)}{\cos(A-C)}$.Using the formula $\cos 2B = \frac{1-	an^2 B}{1+	an^2 B}$ and expanding the right side:$\frac{1-	an^2 B}{1+	an^2 B} = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$.Dividing numerator and denominator by $\cos A \cos C$:$\frac{1-	an^2 B}{1+	an^2 B} = \frac{1-	an A 	an C}{1+	an A 	an C}$.By componendo and dividendo or cross-multiplication:$(1-	an^2 B)(1+	an A 	an C) = (1+	an^2 B)(1-	an A 	an C)$.$1 + 	an A 	an C - 	an^2 B - 	an^2 B 	an A 	an C = 1 - 	an A 	an C + 	an^2 B - 	an^2 B 	an A 	an C$.Simplifying the equation:$2 	an A 	an C = 2 	an^2 B$.$	an^2 B = 	an A 	an C$.Therefore,$	an A, 	an B, 	an C$ are in Geometric Progression $(G.P.)$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

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

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