In a triangle ABC,A,B,C are in G.P with common ratio 2 then 1b+1c−1a is =

We have B=2A,C=4A⇒7A=π⇒A=π/7→1b+1c−1a=12R1sin⁡2A+1sin⁡4A−1sin⁡A=2sin⁡Asin⁡4A+2sin⁡Asin⁡2A−2sin⁡2Asin⁡4A4Rsin⁡Asin⁡2Asin⁡4A=14Rcos⁡2A−cos⁡2A+cos⁡A−cos⁡Asin⁡Asin⁡2Asin⁡4A=0