Prove that tan-1x + cot-1 (x+1) = tan-1 (1 + x + X^2)

Lhs=tan-1x+cot-1x=pi/2

=tan-1x+pi/2-tan-1 (x+1)

now tan-1x-tan-1y=tan-1{ (x-y) / [1-xy]}

so

=tan-1{ (x-x-1) / (1+x+x^2) + pi/2

=tan-1{-1 / (1+x+x^2) }+pi/2

tan is odd func.

so tan-1 (-x) = - tan-1x and tan-1 (1/x) = cot-1x

so

-cot-1 (1+x+x^2) + pi/2

=tan-1 (1+x+x^2)

so

lhs=rhs