If A, B, C are the angles of a triangle such that cot⁡A2=3 tan⁡C2, then sin A, sin B, sin C are in

Given cot⁡A2⋅cot⁡C2=3⇒cos⁡A2⋅cos⁡C2sin⁡A2⋅sin⁡C2=3⇒cos⁡A−C2cos⁡A+C2=2 (using componendo and dividendo)⇒2sin⁡A+C2cos⁡A−C22sin⁡A+C2⋅cos⁡A+C2=2⇒2sin⁡B=sin⁡A+sin⁡C