Question

If in traingle ABC $\cos 2B=\frac{\cos (A+C)}{\cos (A-C)}$, 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. $Noneofthese$

Answer 1

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

In a $△ABC$,

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

$\text{Using componendo - dividendo}$

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

$=\frac{1-\cos 2B}{1+\cos 2B}=\frac{\cos (a-c)-\cos (a+c)}{\cos (a-c)+\cos (a+c)}$

$\Rightarrow \therefore$ $1-\cos 2B=2{\sin }^{2}B$

$1+\cos 2B=2{\cos }^{2}B$

Cos C + Cos D = $2Cos\frac{(C+D)}{2}\cdot \cos \frac{(C-D)}{2}$

$-2\sin \frac{(c+n)}{2}\cdot \sin (\frac{c-p)}{2}$

Now,

$\frac{2{\sin }^{2}B}{2{\cos }^{2}B}$

$-2\sin \frac{(a-c+a+c)}{2}\sin \left(\frac{\alpha -c-a-c}{2}\right)$

_______________________________

$2\cos \left(\frac{a-(+a+1}{2}\right)\cdot \cos \left(\frac{a-(-a-c}{2}\right)$

$\therefore {\tan }^{2}B=\tan A\cdot \tan C$

Option B