Derivative of a quotient involving sine and cosine

Apply the quotient rule: $h'(x) = \dfrac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ with $f(x) = 1+\cos(x)$, so $f'(x) = -\sin(x)$, and $g(x) = 1-\sin(x)$, so $g'(x) = -\cos(x)$.

Substituting gives $h'(x) = \dfrac{-\sin(x)(1-\sin(x)) - (1+\cos(x))(-\cos(x))}{(1-\sin(x))^2}$. Expanding the numerator: $-\sin x + \sin^2 x + \cos x + \cos^2 x = \cos x - \sin x + 1$ (using $\sin^2 x + \cos^2 x = 1$). So $h'(x) = \dfrac{1 + \cos(x) - \sin(x)}{(1-\sin(x))^2}$.