Question 112509
I presume you want to simplify the expression {{{tan^2(alpha)-sin^2(alpha) / tan^2(alpha)sin^2(alpha)}}}.  (I'm using {{{alpha}}} because theta doesn't render).


First thing to do is realize that since {{{tan(alpha)=sin(alpha)/cos(alpha)}}}, {{{tan^2(alpha)=sin^2(alpha)/cos^2(alpha)}}}.  Now we can write:


{{{((sin^2(alpha)/cos^2(alpha))-sin^2(alpha)) / (sin^2(alpha)/cos^2(alpha))sin^2(alpha)}}}.  


{{{((sin^2(alpha)/cos^2(alpha))-sin^2(alpha)/1) / (sin^2(alpha)/cos^2(alpha))sin^2(alpha)}}}.  


{{{((sin^2(alpha)-sin^2(alpha)cos^2(alpha))/cos^2(alpha)) / (sin^4(alpha)/cos^2(alpha))}}}.  


{{{(sin^2(alpha)(1-cos^2(alpha))/cos^2(alpha)) / (sin^4(alpha)/cos^2(alpha))}}}.  Factoring out a {{{sin^2(alpha)}}}


{{{(sin^2(alpha)(sin^2(alpha))/cos^2(alpha)) / (sin^4(alpha)/cos^2(alpha))}}}.  Because {{{sin^2(alpha)+cos^2(alpha)=1}}}, so {{{(1-cos^2(alpha))=sin^2(alpha)}}}


{{{(sin^4(alpha)/cos^2(alpha) )/ (sin^4(alpha)/cos^2(alpha))=1}}}.



Hope that was what you were looking for.

John