Answer
$-\sec x$
Work Step by Step
Using the given identities: $\frac{\csc^2x-\cot^2x}{\sin(-x)\cot x}=\frac{\frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}}{-\sin x\frac{\cos x}{\sin x}}=\frac{\frac{1-\cos^2x}{\sin^2x}}{-\cos x}=\frac{\frac{\sin^2x}{\sin^2x}}{-\cos x}=\frac{1}{-\cos x}=-\sec x$