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If $\cos 2 B-\frac{\cos (A+C)}{\cos (A-C)}$, Then $\tan A, \tan B, \tan C$ are in
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Geometric Progression.
$\begin{aligned} & \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{\cos A \cos C(1-\tan A \tan C)}{\cos A \cos C(1+\tan A \tan C)} \\ & \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{(1-\tan A \tan C)}{(1+\tan A \tan C)}\end{aligned}$
$\begin{aligned} &\left(1-\tan ^2 \mathrm{~B}\right)(1+\tan \mathrm{A} \tan \mathrm{C}) \\ & \quad=(1-\tan \mathrm{A} \tan \mathrm{C})\left(1+\tan ^2 \mathrm{~B}\right) \\ & \quad 1+\tan \mathrm{A} \tan \mathrm{C}-\tan ^2 \mathrm{~B}-\tan ^2 \mathrm{~B} \tan \mathrm{A} \tan \mathrm{C} \\ & \quad=1+\tan ^2 \mathrm{~B}-\tan \mathrm{A} \tan \mathrm{C}-\tan \mathrm{A} \tan \mathrm{C} \tan ^2 \mathrm{~B} \\ & \quad 2 \tan \mathrm{A} \tan \mathrm{C}=2 \tan ^2 \mathrm{~B} \\ & \quad \tan ^2 \mathrm{~B}=\tan \mathrm{A} \cdot \tan \mathrm{C} \\ & \therefore \quad \tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} \text { are in G.P. }\end{aligned}$
$\begin{aligned} &\left(1-\tan ^2 \mathrm{~B}\right)(1+\tan \mathrm{A} \tan \mathrm{C}) \\ & \quad=(1-\tan \mathrm{A} \tan \mathrm{C})\left(1+\tan ^2 \mathrm{~B}\right) \\ & \quad 1+\tan \mathrm{A} \tan \mathrm{C}-\tan ^2 \mathrm{~B}-\tan ^2 \mathrm{~B} \tan \mathrm{A} \tan \mathrm{C} \\ & \quad=1+\tan ^2 \mathrm{~B}-\tan \mathrm{A} \tan \mathrm{C}-\tan \mathrm{A} \tan \mathrm{C} \tan ^2 \mathrm{~B} \\ & \quad 2 \tan \mathrm{A} \tan \mathrm{C}=2 \tan ^2 \mathrm{~B} \\ & \quad \tan ^2 \mathrm{~B}=\tan \mathrm{A} \cdot \tan \mathrm{C} \\ & \therefore \quad \tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C} \text { are in G.P. }\end{aligned}$
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