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

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

4tan^(-1)((1)/(5))=tan^(-1)((1)/(70))+tan^(-1)((1)/(99))+(pi )/(4)

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

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

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

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

Prove that : tan^(-1)(1/2) + tan^(-1)(1/3) = tan^(-1)(3/5) + tan^(-1)(1/4) = pi/4

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

tan^(-1)(n+1)+tan^(-1)(n-1)=tan^(-1)(8/31)

(tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+tan^(-1)((1)/(13))+......+tan^(-1)((1)/(381)))=(m)/(n) where m,n in N, then find least value of (m+n)