If cos⁡2B=cos⁡(A+C)cos⁡(A−C) then tanA, tanB, tanC are in

cos⁡2B1=cos⁡(A+C)cos⁡(A−C)applying componendo and dividendo, we get1−cos⁡2B1+cos⁡2B=cos⁡(A−C)−cos⁡(A+C)cos⁡(A−C)+cos⁡(A+C)or  2sin2⁡B2cos2⁡B=2sin⁡Asin⁡C2cos⁡Acos⁡Cor tan2⁡B=tan⁡Atan⁡CThus, tanA, tanB, tanC are in G.P.