Question 1058012
Multiply the first term by cos(x)/cos(x), and the second term by cos(x + y)/cos(x + y), which gives:


{{{(sin(x + y)/cos(x + y))*(cos(x)/cos(x)) - (sin(x)/cos(x))(cos(x + y)/cos(x + y))}}}


{{{ (sin(x + y)cos(x) - sin(x)cos(x + y)) / cos(x)cos(x + y) }}}


Expanding sin(x + y) and cos(x + y) then gives:


{{{ ((sin(x)cos(y)+cos(x)sin(y))cos(x) - sin(x)(cos(x)cos(y) - sin(x)sin(y))) / cos(x)cos(x + y) }}}


{{{ ((sin(x)cos(y)cos(x) + cos^2(x)sin(y)) - sin(x)cos(x)cos(y) + sin^2(x)sin(y)) / cos(x)cos(x + y) }}}


{{{ (cos^2(x)sin(y) + sin^2(x)sin(y)) / cos(x)cos(x + y) }}}


{{{ ((cos^2(x) + sin^2(x))sin(y)) / cos(x)cos(x + y) }}}


{{{ ((1)sin(y)) / cos(x)cos(x + y) }}}


{{{ sin(y) / cos(x)cos(x + y) }}}


If you want to go a step farther, the above expression is equivalent to:


{{{ sec(x)sin(y) / cos(x + y) }}}