Question

If $\cos 2B = \frac{\cos(A + C)}{\cos(A - C)}$, then $\tan A,\mspace{6mu}\tan B,\mspace{6mu}\tan C$ are in

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

Answer 1

Answer: (B)

G.P.

Answer 2

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

⇒ $\frac{1 - \tan^{2}B}{1 + \tan^{2}B} = \frac{1 - \tan A\tan C}{1 + \tan A\tan C}$

⇒ $1 + \tan^{2}B - \tan A\tan C - \tan A\tan C\tan^{2}B$

$= 1 - \tan^{2}B + \tan A\tan C - \tan A\tan C\tan^{2}B$

⇒ $2\tan^{2}B = 2\tan A\tan C \Rightarrow \tan^{2}B = \tan A\tan C$

Hence, tan A, tan B and $\tan$C will be in G.P.