Official solution
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, \[\cos \beta = {1 \over {\sqrt {{x^2} + 1} }}\]
\[\therefore\] \[{1 \over {\sqrt {{x^2} + 2x + 2} }} = {1 \over {\sqrt {{x^2} + 1} }}\]
\[ \Rightarrow {x^2} + 2x + 2 = {x^2} + 1\]
\[ \Rightarrow 2x + 2 = 1\]
\[ \Rightarrow 2x = - 1\]
\[ \Rightarrow x = -{1 \over 2}\]