## Algebra and Trigonometry 10th Edition

$\frac{sec^2x-1}{sin^2x}=sec^2x$ (e)
Use the Pythagorean Identity: $tan^2x+1=sec^2x$ $tan^2x=sec^2x-1$ And: $tan~x=\frac{sin~x}{cos~x}$ $\frac{sec^2x-1}{sin^2x}=\frac{tan^2x}{\frac{sin^2x}{1}}=\frac{sin^2x}{cos^2x}\frac{1}{sin^2x}=\frac{1}{cos^2x}=sec^2x$