tan^(-1)(1)/(3)+tan^(-1)(2)/(9)+......+tan^(-1)(2^(n-1))/(1+2^(2n-1))+

tan^(-1)(1)/(4)+tan^(-1)(2)/(9)=tan^(-1)(1)/(2)

tan^(-1)((1)/(4))+tan^(-1)((2)/(9))=tan^(-1)((1)/(2))

"tan"^(-1)((1)/(3))+ "tan"^(-1)((2)/(9)) + tan^(-1)((4)/(3^(3))) + ....oo 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 .

tan^(-1)((1)/(1+1.2))+tan^(-1)((1)/(1)+2.3)+...+tan^(-1)((1)/(1+n(n+1)))=tan^(-1)theta then find the value of theta.

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

tan^(- 1)(1/4)+tan^(- 1)(2/9)=1/2tan^(- 1)(4/3)

tan^(- 1)(1/4)+tan^(- 1)(2/9)=1/2tan^(- 1)(4/3)

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