Question 451411
The top is a difference of two squares.  Remember that {{{ a^2-b^2=(a+b)(a-b) }}}  In your case, a is cos(x)+sin(x), and b is cos(x)-sin(x).  Substitute those in and the first parentheses is {{{ (cos(x)+sin(x) + cos(x)-sin(x)) }}}, which simplifies to 2cos(x).  For the other, don't forget to distribute the subtraction: {{{ cos(x)+sin(x)-(cos(x)-sin(x)) }}}, which simplifies to 2sin(x).  Thus, at this point we have: {{{ (2cos(x)*2sin(x))/(2sin(2x)) }}}.

The double angle formula for sine says {{{ sin(2x) = 2sin(x)cos(x) }}}.  The denominator is then {{{ 2*2sin(x)cos(x) }}}.  If we simplify the numerator and denominator, we have {{{ (4sin(x)cos(x))/(4sin(x)cos(x)) }}}, which is, of course, 1 :)