Find the sum `tan^-1 (x/ (1+1.2x^2))+tan^-1 (x/(1+2.3x^2))+…+tan^-1 (x/(1+n(n+1)x^2)) xgt0`

Sum the series tan^-1 (x/(1+1*2x^2))+tan^-1 (x/(1+2*3x^2))+…+tan^-1 (x/(1+n*(n+1)x^2))

Sum the series tan^-1 (x/(1+1*2x^1))+tan^-1 (x/(1+2*3x^2))+…+tan^-1 (x/(1+n*(n+1)x^2))

Using the principle of mathematical induction to show that tan^(−1)(x/(1+1.2.x^2))+tan^(−1)(x/(1+2.3.x^2))+.....+tan^(−1)(x/(1+n(n+1)x^2)) = tan^(-1)(n+1)x-tan^(-1)x , forall x in N .

For all n in N, x in R, tan^(-1) [ ( x)/( 1.2+ x^2) ] + tan^(-1) [ (x)/( 2.3+ x^2) ] + …. + tan^(-1) [ ( x)/( n(n+1) +x^2) ] =

[ If xgt0 then which of the following is true 1) tan^(-1)xgt(x)/(1+x^(2)), 2) tan^(-1)x=(x)/(1+x^(2)) 3) tan^(-1)xlt(x)/(1+x^(2)), 4) tan^(-1)x!=(x)/(1+x^(2))]

AA n in N,x in R,tan^(-1)[(x)/(1.2+x^(2))]+tan^(-1)[(x)/(2.3+x^(2))]+......+tan^(-1)[(x)/(n(n+1)+x^(2))]=

tan^-1 (x-1)/(x-2) + tan^-1 (x+1)/(x+2)=pi/4. find X

Find the sum: tan x + 1/2 tan x/2+ 1/2^(2) tan x/2^(2) +.....+1/2^(n-1) tan x/2^(n-1)

If y = tan^(-1) ((1)/(1 + x + x^(2))) + tan^(-1) ((1) /(x^(2) + 2x + 3)) + tan^(-1) ((1)/(x^(2) + 5x + 7)) + ….n terms , then y(0) is