Here's an excellent example of how one simple trigonometric identity can spare you a significant amount of work on this derivative.

More specifically, you can use the fact that

#color(blue)(sin^x + cos^2x = 1 => sin^2x = 1 - cos^2x)#

to rewrite your function as

#y = (1 - cos^2x)/(1-cosx) = (color(red)(cancel(color(black)((1-cosx)))) * (1 + cosx))/(color(red)(cancel(color(black)((1-cosx))))) = 1 + cosx#

The derivative of #y# will thus be

#y^' = d/dx(1 + cosx) = color(green)(-sinx)#

Now, let's assume that you didn't notice you could use this identity to simplify your function.

Since your function can be written as

#y = f(x)/g(x)#, with #g(x)!=0#

you can use the quotient rule to differentiate it by

#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2#

Using this approach, the derivative of #y# would be

#y^' = ([d/dx(sin^2x)] * (1 - cosx) - sin^2x * d/dx(1-cosx))/(1-cosx)^2#

You can find #d/dx(sin^2x)# by using the chain rule to write

#sin^2x = u^2#, with #u = sinx#

This will get you

#d/dx(u^2) = d/(du)u^2 * d/dx(u)#

#d/dx(u^2) = 2u * d/dx(sinx)#

#d/dx(sin^2x) = 2sinx * cosx#

Take this back to your target derivative to get

#f^' = (2 * sinx * cosx * (1-cosx) - sin^2x * sinx)/(1-cosx)^2#

#f^' = (2sinxcosx - 2sinxcos^2x - sin^3x)/(1-cosx)^2#

You can write this as

#f^' = (-sinx(-2cosx + 2cos^2x + sin^2x))/(1-cosx)^2#

At this point, you could actually use the same identity to write

#f^' = (-sinx * (-2cosx + cos^2x + overbrace(cos^2x + sin^2x)^(color(red)("=1"))))/(1-cosx)^2#

#f^' = (-sinx * (1 - 2cosx + cos^2x))/(1-cosx)^2#

Since you know that

#color(blue)( (a-b)^2 = a^2 - 2ab + b^2)#

you can write

#1 - 2cosx + cos^2x = (1 - cosx)^2#

Finally, plug this back into the expression for #f^'# to get

#f^' = (-sinx * color(red)(cancel(color(black)((1-cosx)^2))))/color(red)(cancel(color(black)((1-cosx)^2))) = color(green)(-sinx)#

So there you have it, the same result, but significantly more work to do.