In order to derive #cot(x)/(1-sin(x))#, you need to apply the quotient rule first.
The quotient rule states this:
#(f'(x)g(x) - g'(x)f(x))/g(x)^2#
Let's assume #(1-sin(x))# is #g(x)#.
Now, the #f(x)# that we will use for the quotient rule is not the equation they give us. It is the numerator, which is #cot(x)#. So let's plug in our newly found #f(x)# and #g(x)#, which is:
#(d/dxcot(x)(1-sin(x))-d/dx(1-sin(x))(cot(x)))/(1-sin(x))^2#
Now, the derivative of #cot(x)# is #-csc^2(x)# and the derivative of #(1-sin(x))# is #-cos(x)#.
After finding the derivatives of #f(x)# and #g(x)#, we rewrite our previous equation:
#(-csc^2(x)(1-sin(x))-(-cos(x))(cot(x)))/(1-sin(x))^2#
Now we split the numerator into two parts: #(cot(x)cos(x))/(1-sin(x))^2-(csc^2(x)(1-sin(x)))/(1-sin(x))^2#
For #(csc^2(x)(1-sin(x)))/(1-sin(x))^2# we can cancel out one #(1-sin(x))# from the numerator and denominator, so this simplifies to: #(csc^2(x))/(1-sin(x))#
So the final answer is: #f'(x) = (cos(x)cot(x))/(1-sin(x))^2-csc^2(x)/(1-sin(x))#