Sinx + sinxcot^2x=cscx .... Prove this identite!?!?!?!

sin(x) + sin(x)·cot²(x)

= sin(x) + sin(x)·[cos²(x)/sin²(x)]        since cot(x) = cos(x) / sin(x)

= sin(x) + cos²(x)/sin(x)                     cancel out the sin(x) factor

= sin²(x)/sin(x) + cos²(x)/sin(x)          write both terms with the common denominator of sin(x)

= [ sin²(x) +  cos²(x) ] sin(x)              add, using the common denominator

=  1 / sin(x)                                      sin²(x) +  cos²(x)  =  1

= csc(x)

$${\mathtt{Sinx}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{sinxcot}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{cscx}} \Rightarrow {\mathtt{sinx}} = {\mathtt{cscx}}{\mathtt{\,-\,}}{{\mathtt{sinxcot}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{x}}$$