the question does not present the options, but this does not interfere with the resolution
we have
sin x/ (1-cos x)
we know that
sin²x+cos²x=1-----------> sin²x=1-cos²x
and
difference of squares
(a+b)*(a-b)=a²-b²
so
The idea is to make the difference of squares (1-cos²x)in denominator.so
multiply the expression by (1+cos x)/(1+cos x)
[sin x/ (1-cos x)]*[(1+cos x)/(1+cos x)]=[sin x*(1+cos x)]/[ (1-cos x)/(1+cos x)]
=[sin x*(1+cos x)]/[ (1-cos²x)]
=[sin x*(1+cos x)]/[ sin²x]
=(1+cos x)/sin x
=(1/sin x)+(cos x/sin x)
=csc x+cot x
therefore
the answer is
the first step is multiply the expression by (1+cos x)/(1+cos x)