Question

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

• A. A.P
• B. H.P
• C. G.P.
• D. A.G.P

Answer 1

Answer: (C)

G.P.

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

Using componendo-dividendo

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

$\frac{2{\cos }^{2}B-1+1}{1-2{\sin }^{2}B-1}=\frac{\cos A\cos C-\sin AsinC+\cos A\cos C+\sin A\sin C}{\cos A\cos C-\sin A\sin C-\cos A\cos C-\sin A\sin C}$

$-{\cot }^{2}B=-\cot A\cot C$

${\tan }^{2}B=\tan A\tan C$
Hence $\tan A,\tan B,\tan C$ are in $G.P$
Hence, option ${}^{\prime }{C}^{\prime }$ is correct.