Question 1136237

here is other way, you might like it better

{{{tan(A)/(1 - cot(A)) + cot(A)/(1 - tan(A))}}}


{{{(sin(A)/cos(A))/(1 - cos(A)/sin(A)) + (cos(A)/sin(A))/(1 - sin(A)/cos(A))}}}



={{{(sin(A)/cos(A))/(1 - cos(A)/sin(A)) + (cos(A)/sin(A))/(1 - sin(A)/cos(A))}}}


={{{((sin(A)/cos(A))/((sin(A) - cos(A))/sin(A)))+ ((cos(A)/sin(A))/((cos(A) - sin(A))/cos(A)))}}}


={{{sin^2(A)/(cos(A)(sin(A) - cos(A)))+ cos^2(A)/(sin(A)(cos(A) - sin(A))))}}}


to have same denominators you can write {{{sin(A)(cos(A) - sin(A))}}} as {{{-sin(A)(sin(A) - cos(A)))}}}



={{{sin^2(A)/(cos(A)(sin(A) - cos(A)))- cos^2(A)/(sin(A)(sin(A) - cos(A))))}}}



={{{(sin^2(A)sin(A) - cos^2(A)cos(A))/(sin(A)cos(A)(sin(A) - cos(A))))}}}



={{{(sin^3(A) - cos^3(A))/(sin(A)cos(A)(sin(A) - cos(A))))}}}


use factoring cube difference rule in numerator


={{{((sin(A) - cos(A))(sin^2(A)+2sin(A)cos(A)+cos^2(A)))/(sin(A)cos(A)(sin(A) - cos(A))))}}}...........simplify


={{{(cross((sin(A) - cos(A)))(sin^2(A)+2sin(A)cos(A)+cos^2(A)))/(sin(A)cos(A)cross((sin(A) - cos(A)))))}}}

={{{(sin^2(A)+2sin(A)cos(A)+cos^2(A))/(sin(A)cos(A))}}}

since {{{sin^2(A)+cos^2(A)=1}}}

={{{(1+2sin(A)cos(A))/(sin(A)cos(A))}}}


={{{1/(sin(A)cos(A))+(2sin(A)cos(A))/(sin(A)cos(A))}}}


={{{1/(sin(A)cos(A))+2}}}......use sec and csc to express sin and cos


={{{1/(sin(A)cos(A))+2}}}.....{{{sin(A)=1/csc(A)}}},{{{cos(A)=1/sec(A)}}}


={{{1/((1/csc(A))(1/sec(A)))+2}}}


={{{1/(1/csc(A)sec(A))+2}}}


={{{csc(A)sec(A)+2}}}