reserve x,x1,x2,x3 for Real;

theorem
  sin(x/2)<>0 & cos(x/2)<>0 & 1-(tan(x/2))^2<>0 implies cosec(x/2)= sqrt
  ((2*sec(x))/(sec(x)-1)) or cosec(x/2)=-sqrt((2*sec(x))/(sec(x)-1))
proof
  assume that
A1: sin(x/2)<>0 and
A2: cos(x/2)<>0 and
A3: 1-(tan(x/2))^2<>0;
  set b=(sec(x/2))^2;
  set a=1-(tan(x/2))^2;
A4: b-a=(1+(tan(x/2))^2)-(1-(tan(x/2))^2) by A2,Th11
    .=2*(tan(x/2))^2;
  sqrt((2*sec(x))/(sec(x)-1)) =sqrt((2*((sec(x/2))^2/(1-(tan(x/2))^2)))/(
  sec(2*(x/2))-1)) by A1,A2,Th13
    .=sqrt((2*((sec(x/2))^2/(1-(tan(x/2))^2))) /(((sec(x/2))^2/(1-(tan(x/2))
  ^2))-1)) by A1,A2,Th13;
  then
A5: sqrt((2*sec(x))/(sec(x)-1))=sqrt(((2*(b/a))*a)/((b/a-1)*a)) by A3,
XCMPLX_1:91
    .=sqrt((2*((b/a)*a))/(b/a*a-1*a))
    .=sqrt((2*((b/a)*a))/(b-a)) by A3,XCMPLX_1:87
    .=sqrt(2*b/(2*(tan(x/2))^2)) by A3,A4,XCMPLX_1:87
    .=sqrt((sec(x/2))^2/(tan(x/2))^2) by XCMPLX_1:91
    .=sqrt((sec(x/2)/(tan(x/2)))^2) by XCMPLX_1:76
    .=sqrt((cosec(x/2))^2) by A2,Th1
    .=|.cosec(x/2).| by COMPLEX1:72;
  per cases;
  suppose
    cosec(x/2)>=0;
    hence thesis by A5,ABSVALUE:def 1;
  end;
  suppose
    cosec(x/2)<0;
    then sqrt((2*sec(x))/(sec(x)-1))*(-1) =(-cosec(x/2))*(-1) by A5,
ABSVALUE:def 1;
    hence thesis;
  end;
end;
