Question 1059832
To make it easier, let {{{ sin(x) = z }}}
Then I have:
{{{ ( z^2 - 2z + 1 ) / ( 1 - z^2 ) }}}
{{{ ( z - 1 )^2 / ( ( 1 + z )*( 1 - z ) ) }}}
{{{ ( z - 1 )^2 / ( ( 1 + z )*( -( z - 1 ) ) ) }}}
{{{ ( z - 1 ) / ( ( 1 + z )*( -1 ) ) }}}
{{{ ( -( z - 1 ) ) / ( 1 + z ) }}}
{{{ ( 1 - z ) / ( 1 + z ) }}}
{{{ ( 1 - sin(x) ) / ( 1 + sin(x) ) }}}
Hope it makes sense