`tan^(-1)n+cot^(-1)(n+1)=tan^(-1)(n^(2)+n+1)`

tan^(-1)n ,tan^(-1)(n+1) and tan^(-1)(n+2),n in N , are the angles of a triangle if n= 1 (b) 2 (c) 3 (d) None of these

tan^(-1)n,tan^(-1)(n+1) and tan^(-1)(n+2),n in N are the angles of a triangle if n= (a) 1(b)2(c)3(d) None of these

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

If S= tan^(-1) ((1)/(n^(2) +n+1)) + tan^(-1) ((1)/(n^(2) + 2n+ 3))+ ….+ tan^(-1) ((1)/(1+(n+19) (n+20))) , then tan (s) is equal to

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 .

If x!=n and cot^(-1)x+cot^(-1)(n^(2)-x+1)=cot^(-1)(n-1) then x=

sum_(n=1)^(n)tan^(-1)((n)/(n))

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

tan^(-1)(m/n)+tan^(-1)((n-m)/(n+m))=?