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

The correct option is B G.P.

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

$\Rightarrow \frac{2{\sin }^{2}B}{2{\cos }^{2}B}=\frac{2\sin A\sin C}{2\cos A\cos C}$

$\Rightarrow {\tan }^{2}B=\tan A\tan C$
$\Rightarrow \tan A,\tan B,\tan C$ are in G.P.