[tan^(-1)((m)/(n))-tan^(-1)((m-n)/(m+n))=],[quad -(pi)/(4)],[quad (pi)/(4)],[(3 pi)/(4)],[(5 pi)/(4)]

Prove that tan^(-1)((m)/(n))-tan^(-1)((m-n)/(m+n))=(pi)/(4).

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

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

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

Show that 'tan ^(-1) (m/n)-tan ^(-1) ((m-n)/(m+n))=pi/4'

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

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

Prove that: tan^(-1)(m/n)+tan^(-1)((n-m)/(n+m))=pi/4

Prove that: tan^(-1)(m/n)+tan^(-1)((n-m)/(n+m))=[pi/4; m^(2)/n^(2) > -1

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