\( \frac { 1 } { 1 + \cot x } - \frac { \sin x } { \sin x - \cos x } \)

To simplify the expression \( \frac { 1 } { 1 + \cot x } - \frac { \sin x } { \sin x - \cos x } \), let's first rewrite \( \cot x \) in terms of sine and cosine. We know that: \[ \cot x = \frac{\cos x}{\sin x}, \] so: \[ 1 + \cot x = 1 + \frac{\cos x}{\sin x} = \frac{\sin x + \cos x}{\sin x}. \] Therefore, \[ \frac{1}{1 + \cot x} = \frac{\sin x}{\sin x + \cos x}. \] Now substitute this back into the original expression: \[ \frac{\sin x}{\sin x + \cos x} - \frac{\sin x}{\sin x - \cos x}. \] Next, we need a common denominator to combine these fractions. The common denominator is \((\sin x + \cos x)(\sin x - \cos x)\). Now we can rewrite the fractions: \[ \frac{\sin x(\sin x - \cos x) - \sin x(\sin x + \cos x)}{(\sin x + \cos x)(\sin x - \cos x)}. \] Calculating the numerator: \[ \sin x(\sin x - \cos x) - \sin x(\sin x + \cos x) = \sin^2 x - \sin x \cos x - \sin^2 x - \sin x \cos x = -2 \sin x \cos x. \] Thus, the expression simplifies to: \[ \frac{-2 \sin x \cos x}{(\sin x + \cos x)(\sin x - \cos x)}. \] Utilizing the double angle identity \( \sin(2x) = 2 \sin x \cos x\), we get: \[ \frac{-\sin(2x)}{(\sin x + \cos x)(\sin x - \cos x)}. \] This gives us the final simplified form: \[ \frac{-\sin(2x)}{(\sin x + \cos x)(\sin x - \cos x)}. \]