34
Evaluate: $\frac{\sec^2 45^\circ - \tan^2 45^\circ}{\sin^2 45^\circ}$
Show SolutionHide Solution↓
$\frac{\sec^2 45^\circ - \tan^2 45^\circ}{\sin^2 45^\circ} = \frac{(\sqrt{2})^2-(1)^2}{\left(\frac{1}{\sqrt{2}}\right)^2} = \frac{2-1}{\frac{1}{2}} = 2$