For any positive integer n, define  fn:0,∞→ℝ as fnx=∑j=1ntan−111+x+jx+j−1  for all x∈0,∞  (Here, the inverse trigonometric function tan−1x assume values in −π2,π2   Then, which of the following statement(s) is (are) TRUE?

fn(x)=∑j=1ntan-1(x+j)-(x+j-1)1+(x+j)(x+j-1)fn(x)=∑j=1ntan-1(x+j)-tan-1(x+j-1)fn(x)=tan-1(x+n)-tan-1x∴  tanfn(x)=tantan-1(x+n)-tan-1xtanfn(x)=(x+n)-x1+x(x+n)  tanfn(x)=n1+x2+nx∴sec2fn(x)=1+tan2fn(x)limx→∞sec2fn(x)=limx→∞1+n1+x2+nx2=1