Question 139752
I'm presuming that you mean {{{(sin(x)/csc^2(x))-(cos^2(x)/sec(x))=-cos^3(x)}}} rather than {{{(sin(x)/csc(2x))-(cos(2x)/sec(x))=-cos(3x)}}}, because the second one is a horror I have no intention of spending the rest of the afternoon solving.


{{{sin(x)=1/csc(x)}}} (read is identical to), so {{{1/csc^2(x)=sin^2(x)}}} and 
{{{cos(x)=1/sec(x)}}}


So:
{{{(sin(x)/csc^2(x))-(cos^2(x)/sec(x))=sin^3(x)-cos^3(x)<>-cos^3(x)}}}, except where {{{sin(x)=0}}}