"We can analyze as follows:"
\qquad \qquad \qquad sin[ cot^{-1}( 1 + x ) ] \ = \ cos[ tan^{-1}( x ) ].
"Let:" \qquad \qquad \qquad A = cot^{-1}( 1 + x ) \qquad "and" \qquad B = tan^{-1}( x ).
"Then we have:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad sin[ A ] \ = \ cos[ B ].
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\quad 1/csc[ A ] \ = \ 1/sec[ B ].
\qquad \qquad \qquad \qquad \qquad \qquad \quad 1/sqrt{ 1 + cot^2 A } \ = \ 1/sqrt{ 1 + tan^2 A }.
\qquad 1/sqrt{ 1 + cot^2 [ cot^{-1}( 1 + x ) ] } \ = \ 1/sqrt{ 1 + tan^2 [ tan^{-1}( x ) ] }.
1/sqrt{ 1 + ( cot[ cot^{-1}( 1 + x ) ] )^2 } \ = \ 1/sqrt{ 1 + ( tan[ tan^{-1}( x ) ] )^2 ].
\qquad \qquad \qquad \qquad \qquad \quad 1/sqrt{ 1 + ( 1 + x )^2 } \ = \ 1/sqrt{ 1 + x^2 ].
\qquad \qquad \qquad \qquad \qquad \quad \ sqrt{ 1 + ( 1 + x )^2 } \ = \ sqrt{ 1 + x^2 ].
\qquad \qquad \qquad \qquad \qquad \qquad \quad \ 1 + ( 1 + x )^2 \ = \ 1 + x^2.
\qquad \qquad \qquad \qquad \ 1 + ( 1 + 2 x + x^2 ) \ = \ 1 + x^2.
\qquad \qquad \qquad \qquad \qquad \qquad \quad 2 + 2 x + color{red}cancel{ x^2 } \ = \ 1 + color{red}cancel{ x^2 }.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 2 + 2 x \ = \ 1 .
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 2 x \ = \ -1 .
qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ x \ = \ -1/2 .
"This is our solution !!"
"So, summarizing:"
\qquad "The solution set of:" \qquad \quad sin[ cot^{-1}( 1 + x ) ] \ = \ cos[ tan^{-1}( x ) ]
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ "is:" qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ x \ = \ -1/2 \quad. \qquad \qquad \ square
