Question

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

• A. $AP$
• B. $GP$
• C. $HP$
• D. None of these

Answer 1

The correct option is B

$GP$

Finding the relation in $\tan A,\tan B,\tan C$ :

Given,

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

$\Rightarrow$$\frac{(1–{\tan }^{2}B)}{(1+{\tan }^{2}B)}=\frac{[\cos A\cos C–\sin A\sin C]}{[\cos A\cos C+\sin A\sin C]}$

Dividing the numerator and denominator of RHS by $\cos A\cos C$

$\frac{(1–{\tan }^{2}B)}{(1+{\tan }^{2}B)}=\frac{[1–\tan A\tan C]}{[1+\tan A\tan C]}$

$\Rightarrow$ $\frac{(1–{\tan }^{2}B)}{(1+\tan A\tan C)}=(1+{\tan }^{2}B)(1–\tan A\tan C)$

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

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

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

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

We know that if $a,b,c$ are G.P, then$b^2=ac$.

$\therefore \tan A,\tan B,\tan C$ are in $\mathit{G}\mathbf{.}\mathit{P}\mathbf{.}$

Hence, option ‘B’ is correct.