If $\cos 2 B-\frac{\cos (A+C)}{\cos (A-C)}$, Then $ an A, an B, an C$ are in

Question

If $\cos 2 B-\frac{\cos (A+C)}{\cos (A-C)}$, Then $	an A, 	an B, 	an C$ are in

Aicharya · Accepted Answer

$\begin{aligned} & \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)} \ & \frac{1-	an ^2 B}{1+	an ^2 B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \ & \frac{1-	an ^2 B}{1+	an ^2 B}=\frac{\cos A \cos C(1-	an A 	an C)}{\cos A \cos C(1+	an A 	an C)} \ & \frac{1-	an ^2 B}{1+	an ^2 B}=\frac{(1-	an A 	an C)}{(1+	an A 	an C)}\end{aligned}$ $\begin{aligned} &\left(1-	an ^2 \mathrm{~B}ight)(1+	an \mathrm{A} 	an \mathrm{C}) \ & \quad=(1-	an \mathrm{A} 	an \mathrm{C})\left(1+	an ^2 \mathrm{~B}ight) \ & \quad 1+	an \mathrm{A} 	an \mathrm{C}-	an ^2 \mathrm{~B}-	an ^2 \mathrm{~B} 	an \mathrm{A} 	an \mathrm{C} \ & \quad=1+	an ^2 \mathrm{~B}-	an \mathrm{A} 	an \mathrm{C}-	an \mathrm{A} 	an \mathrm{C} 	an ^2 \mathrm{~B} \ & \quad 2 	an \mathrm{A} 	an \mathrm{C}=2 	an ^2 \mathrm{~B} \ & \quad 	an ^2 \mathrm{~B}=	an \mathrm{A} \cdot 	an \mathrm{C} \ & 	herefore \quad 	an \mathrm{A}, 	an \mathrm{B}, 	an \mathrm{C} 	ext { are in G.P. }\end{aligned}$

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

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