## Algebra and Trigonometry 10th Edition

The identity is verified. $\frac{sec^2θ-1}{sec^2θ}=sin^2θ$
$sec~θ=\frac{hyp}{adj}=\frac{c}{b}$ $sin~θ=\frac{opp}{hyp}=\frac{a}{c}$ Use the Pythagorean Identity: $a^2+b^2=c^2$ $a^2=c^2-b^2$ $\frac{sec^2θ-1}{sec^2θ}=\frac{(\frac{c}{b})^2-1}{(\frac{c}{b})^2}=\frac{\frac{c^2-b^2}{b^2}}{\frac{c^2}{b^2}}=\frac{c^2-b^2}{b^2}~\frac{b^2}{c^2}=\frac{c^2-b^2}{c^2}=\frac{a^2}{c^2}=sin^2θ$