## Trigonometry (11th Edition) Clone

The following statement is an identity: $1+cot^2~x = \frac{sec^2~x}{sec^2~x-1}$
We can see that the graph of $~~y = 1+cot^2~x~~$ is the same as the graph of $~~y = \frac{sec^2~x}{sec^2~x-1}$ Therefore, the following statement is an identity: $1+cot^2~x = \frac{sec^2~x}{sec^2~x-1}$