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  4. If cos 2 B=( cos (A+C)/ cos (A-C)), then tan A, tan B, tan C are in

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

Trigonometric Functions

Solution:

$\frac{\cos 2 B}{1}=\frac{\cos (A+C)}{\cos (A-C)}$
Applying componendo and dividendo, we get
$\frac{1-\cos 2 B}{1+\cos 2 B}=\frac{\cos (A-C)-\cos (A+C)}{\cos (A-C)+\cos (A+C)}$
or $\frac{2 \sin ^{2} B}{2 \cos ^{2} B}=\frac{2 \sin A \sin C}{2 \cos A \cos C}$
or $\tan ^{2} B=\tan A \tan C$
Thus, $\tan A, \tan B, \tan C$ are in $G.P$.