Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Ans. $True$. We have $\frac{\cos 2B}{1}=\frac{\cos (A+C)}{\cos (A-C)}$.
Applying componendo and dividend,
$\frac{1-\cos 2B}{1+\cos 2B}=\frac{\cos (A-C)-\cos (A+C)}{\cos (A-C)+\cos (A+C)}$
$\Rightarrow \frac{2{\sin }^{2}B}{2{\cos }^{2}B}=\frac{2\sin A\sin C}{2\cos A\cos C}$
$\Rightarrow {\tan }^{2}B=\tan A\tan C\Rightarrow \tan A,\tan B,\tan C$ are in $G.P.$