## Shaydenparis 2 years ago prove sec^2θ/over/sec^2θ-1=csc^2

tan²θ + 1 = sec²θ tan²θ = sec²θ - 1 So $$\dfrac{\sec^2 \theta}{\sec^2 \theta - 1} = \dfrac{\sec^2 \theta}{\tan^2 \theta} = \dfrac{\dfrac{1}{\cos^2 \theta}}{\dfrac{\sin^2 \theta}{\cos^2 \theta}} = \dfrac{1}{\sin^2 \theta} = \boxed{\csc^2 \theta}$$ Is this clear?