Question 1178040
 Simplify: 



{{{ (cos(theta)+sin(theta))/(cos(theta)-sin(theta))-(cos(theta)-sin(theta))/(cos(theta)+sin(theta))}}}

 and express the answer in terms of {{{tan(theta)}}}


{{{ (cos(theta)+sin(theta))/(cos(theta)-sin(theta))-(cos(theta)-sin(theta))/(cos(theta)+sin(theta))}}}..........common denominator{{{ (cos(theta)-sin(theta))(cos(theta)+sin(theta))=(cos^2(theta)-sin^2(theta))}}}


{{{((cos(theta)+sin(theta))(cos(theta)+sin(theta)) -(cos(theta)-sin(theta))(cos(theta)-sin(theta)))/(cos^2(theta)-sin^2(theta))}}}


{{{((cos(theta)+sin(theta))^2 -(cos(theta)-sin(theta))^2)/(cos^2(theta)-sin^2(theta))}}}


{{{((cos^2(theta)+2cos(theta)sin(theta)+sin^2(theta))-(cos^2(theta)-2cos(theta)sin(theta)+sin^2(theta)))/(cos^2(theta)-sin^2(theta))}}}


{{{(cos^2(theta)+2cos(theta)sin(theta)+sin^2(theta)-cos^2(theta)+2cos(theta)sin(theta)-sin^2(theta))/(cos^2(theta)-sin^2(theta))}}}


{{{(4cos(theta)sin(theta))/(cos^2(theta)-sin^2(theta))}}}.....use identity {{{cos^2(theta)=1-sin^2(theta)}}}


{{{(4cos(theta)sin(theta))/(1-sin^2(theta)-sin^2(theta))}}}


{{{(4cos(theta)sin(theta))/(1-2sin^2(theta))}}}........Double-Angle Identities {{{ 2cos(theta)sin(theta)=sin(2theta)}}} and {{{1-2sin^2(theta)=cos(2theta) }}}


{{{(2sin(2theta))/cos(2theta) }}}


{{{2tan(2theta)}}}