Question

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

• A. A.P
• B. G,P
• C. H.P
• D. AGP

Answer 1

Answer: (C)

H.P

$tanB=\frac{2sinAsinC}{sin(A+C)}\Rightarrow \frac{sinB}{cosB}=\frac{2sinAsinC}{sinAcosC+cosAsinC}\Rightarrow sinAsinBcosC+cosAsinBsinC=2sinAcosBsinC$
Dividing both sides by $sinAsinBsinC$, we get
$\frac{sinAsinBcosC}{sinAsinBsinC}+\frac{cosAsinBsinC}{sinAsinBsinC}=\frac{2sinAcosBsinC}{sinAsinBsinC}\Rightarrow \frac{cosC}{sinC}+\frac{cosA}{sinA}=\frac{2cosB}{sinB}\Rightarrow \frac{1}{tanC}+\frac{1}{tanA}=\frac{2}{tanB}$
Hence $tanA,tanB,tanC$ are in H.P