#### Answer

The following statement is an identity:
$1+cot^2~x = \frac{sec^2~x}{sec^2~x-1}$

#### Work Step by Step

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:
$1+cot^2~x = \frac{sec^2~x}{sec^2~x-1}$