`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

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

tan^(-1)((1)/(2))+tan^(-1)((1)/(3))=(pi)/(4)|0tan(1)+tan(1)

Prove that: "tan"^(-1)(m)/(n)-tan^(-1)((m-n)/(m+n))=(pi)/(4). m, n gt 0

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

tan^(-1)((a+x)/a)+tan^(-1)((a-n)/a)=(pi)/(6)

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

The value of tan^(-1)((1)/(3))+tan^(-1)((2)/(9))+tan^(-1)((4)/(33))+tan^(-1)((8)/(129))+...n terms is:

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))