The value of limx→0{Sin2(π2−ax)}sec2π2−bx is equal to

Given limit is of the form 1∞∴​ Given  limit =elimx→0 sec2(π2−bx)(sin2(π2−ax)−1) =elimx→0-cos2(π2−ax)cos2(π2−bx) Use L-Hospital rule =elimx→0-sin(2π2−ax)sin(2π2−bx).ab Again use L-Hospital rule = =e−a2b2