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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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Verified Answer
The correct answer is:
G.P.
$\begin{aligned}
& \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\
& \Rightarrow \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{1-\tan A \tan C}{1+\tan A \tan C} \\
& \Rightarrow 1+\tan ^2 B-\tan A \tan C-\tan A \tan C \tan ^2 B \\
&= 1-\tan ^2 B+\tan A \tan C-\tan A \tan C \tan ^2 B \\
& \Rightarrow 2 \tan ^2 B=2 \tan A \tan C \Rightarrow \tan ^2 B=\tan A \tan C
\end{aligned}$
Hence, $\tan A, \tan B$ and $\tan C$ will be in G.P.
& \cos 2 B=\frac{\cos (A+C)}{\cos (A-C)}=\frac{\cos A \cos C-\sin A \sin C}{\cos A \cos C+\sin A \sin C} \\
& \Rightarrow \frac{1-\tan ^2 B}{1+\tan ^2 B}=\frac{1-\tan A \tan C}{1+\tan A \tan C} \\
& \Rightarrow 1+\tan ^2 B-\tan A \tan C-\tan A \tan C \tan ^2 B \\
&= 1-\tan ^2 B+\tan A \tan C-\tan A \tan C \tan ^2 B \\
& \Rightarrow 2 \tan ^2 B=2 \tan A \tan C \Rightarrow \tan ^2 B=\tan A \tan C
\end{aligned}$
Hence, $\tan A, \tan B$ and $\tan C$ will be in G.P.
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