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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
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