Inspection of
#(sin^2(-x) - sin(-x))/(1 - sin(-x)#
reveals that the numerator may be factorised by extracting the term #sin(-x)#.
That is, the expression may be written
#(sin(-x) (sin(-x) - 1))/(1 - sin(-x)#
#sin(-x)# is an example of an odd function, for which #f(-x) = - f(x)# (which means it has rotational symmetry around the origin, compared with even functions, for which #f(-x) = f(x)#, which have reflective symmetry in the y axis)
So, substituting #-sin(x)# for the #sin(-x)# in the term outside the brackets (but leaving the others for now for reasons that will become clear)
#(sin(-x) (sin(-x) - 1))/(1 - sin(-x)#
implies
#(- sin(x) (sin(-x) - 1))/(1 - sin(-x)#
which in turn implies
#(sin(x) (1 - sin(-x)))/(1 - sin(-x))#
The term #(1 - sin(-x))# is a factor of both numerator and denominator so it may be cancelled to leave
#sin(x)#
