" ii) "tan^(-1)(n)/(n+1)-tan^(-1)(2n+1)=(3 pi)/(2)

tan^(-1)((n)/(n+1))-tan^(-1)(2n+1)=(3 pi)/(4)

tan^(-1)((3)/(n))+tan^(-1)((4)/(n))=(pi)/(2)

For n in N ,if tan^(-1)((1)/(3))+tan^(-1)((1)/(4))+tan^(-1)((1)/(5))+tan^(-1)((1)/(n))=(pi)/(4) ,then (n-2)/(15) is equal to

sum_(n=1)^(oo)(tan^(-1)((4n)/(n^(4)-2n^(2)+2))) is equal to ( A) tan^(-1)(2)+tan^(-1)(3)(B)4tan^(-1)(1)(C)(pi)/(2)(D)sec^(-1)(-sqrt(2))

tan^(-1)((n-5)/(n-6))+tan^(-1)((n+5)/(n+6))=(pi)/(4)

Prove that: tan^(-1)((m)/(n))+tan^(-1)((n-m)/(n+m))=[(pi)/(4)(m)/(n)>;-1(-3 pi)/(4)(m)/(n)<-1

sum_(r=1)^(n) tan^(-1)(2^(r-1)/(1+2^(2r-1))) is equal to a) tan^(-1)(2^n) b) tan^(-1)(2)^n-pi/4 c) tan^(-1)(2^(n+1)) d) tan^(-1)(2^(n+1))-pi/4

The nth term of the coresponding series of int_(0)^(1)tan^(-1)xdx is (A) (pi)/(4n) (B) ((1)/(n))tan^(-1)(n-1)(C)(pi)/(2n)(D)tan^(-1)(n)

Show by mathematical induction that tan^(-1) .(1)/(3) + tan^(-1). (1)/(7) + …+tan^(-1). (1)/(n^2 + n +1)= tan^(-1). (n)/(n+2), AA n inN .

Statement 1: If agt0,bgt0, tan^(-1)(a/x)+tan^(-1)(b/x)=(pi)/2 . implies x=sqrt(ab) Statement 2: If m,n epsilonN,ngem, then "tan"^(-1)(m/n)+tan^(-1)(n-m)/(n+m)=(pi)/4 .