## Precalculus (6th Edition)

$\frac{\sin^2\theta}{\cos\theta}=\sec\theta-\cos\theta$
Start with the right side: $\sec\theta-\cos\theta$ Write in terms of sine and cosine: $=\frac{1}{\cos\theta}-\cos\theta$ Get a common denominator: $=\frac{1}{\cos\theta}-\frac{\cos^2\theta}{\cos\theta}$ Simplify: $=\frac{1-\cos^2\theta}{\cos\theta}$ $=\frac{\sin^2\theta}{\cos\theta}$ Since this equals the left side, the identity has been proven.