Given:
The trigonometric expression is,
[tex]\frac{\sec ^2\theta-1}{\sec ^2\theta}[/tex]
Explanation:
Simplify the expression.
[tex]\begin{gathered} \frac{\sec^2\theta-1}{\sec^2\theta}=\frac{\tan ^2\theta}{\sec ^2\theta}\text{ \lbrack}sec^2\theta+tan^2\theta=1\text{\rbrack} \\ =\frac{\sin^2\theta}{\cos^2\theta}\cdot\frac{1}{\sec ^2\theta} \\ =\frac{\sin^2\theta}{\cos^2\theta}\cdot\cos ^2\theta \\ =\sin ^2\theta \end{gathered}[/tex]
So equivalent expression is,
[tex]\sin ^2\theta[/tex]
