Question

prove sec^2θ/over/sec^2θ-1=csc^2

Answer 1

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?