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, $$\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}$$

$$\therefore$$ $${1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}$$

$$ \Rightarrow {x^2} + 2x + 2 = {x^2} + 1$$

$$ \Rightarrow 2x + 2 = 1$$

$$ \Rightarrow 2x = - 1$$

$$ \Rightarrow x = -{1 \over 2}$$