If cos2B=cos(A+C)cos(A-C) then tanA,tanB,tanCare in
AP
GP
HP
None of these
Finding the relation in tanA,tanB,tanC :
Given,
cos2B=cos(A+C)cos(A–C)
⇒(1–tan2B)(1+tan2B)=[cosAcosC–sinAsinC][cosAcosC+sinAsinC]
Dividing the numerator and denominator of RHS by cosAcosC
(1–tan2B)(1+tan2B)=[1–tanAtanC][1+tanAtanC]
⇒ (1–tan2B)(1+tanAtanC)=(1+tan2B)(1–tanAtanC)
⇒1+tanAtanC–tan2B–tanAtan2BtanC=1–tanAtanC+tan2B–tanAtan2B
⇒ tanCtanAtanC+tanAtanC=tan2B+tan2B
⇒ ,2tan2B=2tanAtanC
⇒ tan2B=tanAtanC
We know that if a,b,c are G.P, thenb2=ac.
∴tanA,tanB,tanC are in G.P.
Hence, option ‘B’ is correct.