limx→π2 1−tan⁡x2(1−sin⁡x)1+tan⁡x2(π−2x)3 is equal to

put, x=π2−h as x→π2,h→0Given, limit =limh→0 1−tan⁡π4+h2(1−cos⁡h)1+tan⁡π2+h2(2h)3=limh→0 tan⁡h2limh→0 2⋅sin2⁡h28h3=limh→0 14tan⁡h22×h2×limh→0 sin⁡h2h22×14=132