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Shaydenparis
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?