Question

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

• A. A.P.
• B. G.P.
• C. H.P.
• D. none of these

Answer 1

Answer: (B)

$\frac{\cos 2B}{1}=\frac{\cos (A+C)}{\cos (A-C)}$
Applying componendo and dividendo, we get
$\frac{1-\cos 2B}{1+\cos 2B}=\frac{\cos (A-C)-\cos (A+C)}{\cos (A-C)+\cos (A+C)}$
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 $G.P$.