reserve x,x1,x2,x3 for Real;

theorem Th8:
  cos(x)<>0 implies cos(2*x)=(1-(tan(x))^2)/(1+(tan(x))^2)
proof
  assume cos(x)<>0;
  then
A1: (cos(x))^2<>0 by SQUARE_1:12;
  cos(2*x) = ((cos(x))^2-(sin(x))^2)*1 by Th7
    .=((cos(x))^2-(sin(x))^2)*((cos(x))^2/(cos(x))^2) by A1,XCMPLX_1:60
    .=((cos(x))^2-(sin(x))^2)/(cos(x))^2*(cos(x))^2 by XCMPLX_1:75
    .=((cos(x))^2/(cos(x))^2-(sin(x))^2/(cos(x))^2)*(cos(x))^2 by XCMPLX_1:120
    .=(1-(sin(x))^2/(cos(x))^2)*(cos(x))^2 by A1,XCMPLX_1:60
    .=(1-(tan(x))^2)*(cos(x))^2/1 by XCMPLX_1:76
    .=(1-(tan(x))^2)*(cos(x))^2/((cos(x))^2+(sin(x))^2) by SIN_COS:29
    .=(1-(tan(x))^2)/(((cos(x))^2+(sin(x))^2)/(cos(x))^2) by XCMPLX_1:77
    .=(1-(tan(x))^2)/(((cos(x))^2/(cos(x))^2+(sin(x))^2/(cos(x))^2)) by
XCMPLX_1:62
    .=(1-(tan(x))^2)/((1+(sin(x))^2/(cos(x))^2)) by A1,XCMPLX_1:60
    .=(1-(tan(x))^2)/((1+(tan(x))^2)) by XCMPLX_1:76;
  hence thesis;
end;
