sin(180-x)-cotxsin(x-90)=cscx
$\displaystyle sin(a \pm b) = sin(a)~cos(b) \pm sin(b)~cos(a)$
So:
$\displaystyle sin(180 - x) = sin(180)~cos(x) - sin(x) ~cos(180) = sin(x)$
and
$\displaystyle sin(x - 90) = sin(x)~cos(90) - sin(90)~cos(x) = -cos(x)$
So your left hand side becomes:
$\displaystyle sin(x) + cot(x)~cos(x)$
Can you take it from here?
-Dan