Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.5 Write Trigonometric Functions and Models - 14.5 Exercises - Quiz for Lessons 14.3-14.5 - Page 945: 5

$\frac{\sin x }{\sin^2 x-1}$

Using the given identities: $\frac{\tan(\pi/2-x)\sec x}{1-\csc^2 x}=\frac{\cot x \frac{1}{\cos x}}{1-\frac{1}{\sin^2 x}}=\frac{\frac{\cos x}{\sin x} \frac{1}{\cos x}}{\frac{\sin^2 x}{\sin^2 x}-\frac{1}{\sin^2 x}}=\frac{\frac{1}{\sin x} }{\frac{\sin^2 x-1}{\sin^2 x}}=\frac{\sin x }{\sin^2 x-1}$