reserve n,m for Element of NAT;
reserve h,k,r,r1,r2,x,x0,x1,x2,x3 for Real;
reserve f,f1,f2 for Function of REAL,REAL;
reserve S for Seq_Sequence;

theorem
  (for x holds f.x=(tan(#)tan).x) & x in dom tan & x-h in dom tan
  implies bD(f,h).x = -(1/2)*(cos(2*x)-cos(2*(h-x)))/((cos(x)*cos(x-h))^2)
proof
  assume that
A1:for x holds f.x=(tan(#)tan).x and
A2:x in dom tan & x-h in dom tan;
A3:cos(x)<>0 & cos(x-h)<>0 by A2,FDIFF_8:1;
  bD(f,h).x = f.x - f.(x-h) by DIFF_1:4
    .= (tan(#)tan).x - f.(x-h) by A1
    .= (tan(#)tan).x - (tan(#)tan).(x-h) by A1
    .= tan.x*tan.x-(tan(#)tan).(x-h) by VALUED_1:5
    .= tan.x*tan.x-tan.(x-h)*tan.(x-h) by VALUED_1:5
    .= sin.x*(cos.x)"*tan.x-tan.(x-h)*tan.(x-h) by A2,RFUNCT_1:def 1
    .= sin.x*(cos.x)"*(sin.x*(cos.x)")-tan.(x-h)*tan.(x-h)
                                                     by A2,RFUNCT_1:def 1
    .= sin.x*(cos.x)"*(sin.x*(cos.x)")
       -(sin.(x-h)*(cos.(x-h))")*tan.(x-h) by A2,RFUNCT_1:def 1
    .= (tan(x))^2-(tan(x-h))^2 by A2,RFUNCT_1:def 1
    .= (tan(x)-tan(x-h))*(tan(x)+tan(x-h))
    .= (sin(x-(x-h))/(cos(x)*cos(x-h)))*(tan(x)+tan(x-h)) by A3,SIN_COS4:20
    .= (sin(h)/(cos(x)*cos(x-h)))*(sin(x+(x-h))/(cos(x)*cos(x-h)))
                                                         by A3,SIN_COS4:19
    .= (sin(h)*sin(2*x-h))/((cos(x)*cos(x-h))^2) by XCMPLX_1:76
    .= (-(1/2)*(cos(h+(2*x-h))-cos(h-(2*x-h))))/((cos(x)*cos(x-h))^2)
                                                          by SIN_COS4:29
    .= -(1/2)*(cos(2*x)-cos(2*(h-x)))/((cos(x)*cos(x-h))^2);
  hence thesis;
end;
