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