For which value of x, sin[cot^-1(x+1)]=cos(tan^-1x).

$\sin[\cot^{-1}(x+1)]=\sin[\sin^{-1}\frac{1}{\sqrt{(x+1)^2+1}}]=\frac{1}{\sqrt{x^2+2x+2}};\cos(\tan^{-1}x)=\cos[\cos^{-1}\frac{1}{\sqrt{1+x^2}}]=\frac{1}{\sqrt{1+x^2}}$

We have $x^2+2x+2=1+x^2$ or $x=-\frac{1}{2}$