Let, ${\cot ^{ - 1}}(1 + x) = \alpha$ $\Rightarrow 1 + x = \cot \alpha$ Now, let ${\tan ^{ - 1}}(x) = \beta$ $\Rightarrow x = \tan \beta$ Given, $\sin ({\cot ^{ - 1}}(x)) = \cos ({\tan ^{ - 1}}(x))$ $\Rightarrow \sin \alpha = \cos \beta$ From $\Delta$ABC, $\sin \alpha = {1 \over {\sqrt {1 + {x^2} + 2x + 1} }}$ $= {1 \over {\sqrt {{x^2} + 2x + 2} }}$ From $\Delta$MNO, $\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}$ $\therefore$ ${1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}$ $\Rightarrow {x^2} + 2x + 2 = {x^2} + 1$ $\Rightarrow 2x + 2 = 1$ $\Rightarrow 2x = - 1$ $\Rightarrow x = -{1 \over 2}$