Given that (1+1+x)tan⁡y=1+1−x  Then sin4y is equal to

tan⁡y=1+1−x1+1+xLet x= cos θ, then1−x=2sin⁡(θ/2);1+x=2cos⁡(θ/2)⇒ tan⁡y=212+sin⁡θ2212+cos⁡θ2=sin⁡π4+sin⁡θ2cos⁡π4+cos⁡θ2=2sin⁡π8+θ4cos⁡π8−θ42cos⁡π8+θ4cos⁡π8−θ4=tan⁡π8+θ4⇒ 4y=π2+θ or  sin⁡4y=cos⁡θ=x