If y=(1+tan⁡A)(1−tan⁡B), where A−B=π4, then (y+1)y+1 is equal to

A−B=π4 or tan⁡(A−B)=tan⁡π4or   tan⁡A−tan⁡B1+tan⁡Atan⁡B=1or   tan⁡A−tan⁡B−tan⁡Atan⁡B=1or   tan⁡A−tan⁡B−tan⁡Atan⁡B+1=2or   (1+tan⁡A)(1−tan⁡B)=2⇒y=2Hence, (y+1)y+1=(2+1)2+1=(3)3=27.