Given the identity:
[tex]\frac{sec^2x}{sec^2x-1}=\csc ^2x[/tex]
Reall from trigonometric identities:
[tex]\text{sec}^2x-1=\tan ^2x[/tex]
Therefore:
[tex]\begin{gathered} \frac{sec^2x}{sec^2x-1}=\frac{sec^2x}{\tan^2x} \\ \implies\frac{sec^2x}{\tan^2x}=\frac{1}{\cos^2x}\div\frac{\sin^2x}{\cos^2x}\text{ where }\begin{cases}\sec ^2x=\frac{1}{\cos^2x} \\ \tan ^2x=\frac{\sin^2x}{\cos^2x}\end{cases} \end{gathered}[/tex]
Change the division sign to multiplication.
[tex]\begin{gathered} \frac{1}{\cos^2x}\times\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin ^2x} \\ \text{The inverse of sine is cosecant.} \\ \implies\frac{sec^2x}{sec^2x-1}=\csc ^2x \end{gathered}[/tex]
Thus, the identity is proved.
