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