If A,B,C are the angles of a triangle and  cos⁡B+cos⁡C=4sin2A2, then tan⁡B2tan⁡C2 is equal to

cos⁡B+cos⁡C=4sin2⁡(A/2)⇒ 2cos⁡B+C2cos⁡B−C2=4sin⁡A2sin⁡π2−B+C2⇒ 2sin⁡A2cos⁡B−C2=4sin⁡A2cos⁡B+C2⇒ cos⁡B2cos⁡C2+sin⁡B2sin⁡C2=2cos⁡B2cos⁡C2−sin⁡B2sin⁡C2⇒3sin⁡B2sin⁡C2=cos⁡B2cos⁡C2⇒tan⁡B2tan⁡C2=13.