Question

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

• A. $\tan A,\tan B,\tan C$ are in A.P.
• B. $\tan A,\tan B,\tan C$ are in G.P.
• C. $\tan A,\tan B,\tan C$ are in H.P.
• D. None of these.

Answer 1

The correct option is B $\tan A,\tan B,\tan C$ are in G.P.

$\cos 2B=\frac{\cos (A+C)}{\cos (A-C)}$

Applying Componendo and Dividendo rule

$⟹\frac{1+\cos 2B}{1-\cos 2B}=\frac{\cos (A-C)+\cos (A+C)}{\cos (A-C)-\cos (A+C)}$

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

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

$⟹{\tan }^{2}B=\tan A\tan C$

So $\tan A,\tan B,\tan C$ are in $GP$