Answer
See proof
Work Step by Step
$$\sec2x=\frac{\cot^2x+1}{\cot^2x-1}$$
$\text{Solution:}$
\begin{align*}
\sec2x&=\frac{1}{\cos2x} ~~~~~\text{Reciprocal Identity}\\
&=\frac{1}{\cos(x+x)} ~~~~~\because 2x=x+x\\
&=\frac{1}{\cos x \cos x-\sin x \sin x } ~~~\text{Cosine Sum Identity}\\
&=\frac{1}{\cos x \cos x-\sin^2 x } ~~~\because \sin x \sin x =\sin^2 x\\
&=\frac{1}{\cos ^2x -\sin^2 x } ~~~~~~\because \cos x \cos x=\cos ^2x\\
&=\frac{1}{\left(\frac{\cos^2 x}{\sin^2x}-1\right)\sin^2x} ~~~~ \text{Factor out} ~\sin^2x \\
&=\frac{1}{(\cot^2x-1)\sin^2x} ~~~~~\because \cot^2x=\frac{\cos^2 x}{\sin^2x}\\
&=\frac{1}{(\cot^2x-1)\frac{1}{\csc^2x}} ~~\because \sin x=\frac{1}{\csc x}\\
&=\frac{\csc^2x}{\cot^2x-1} ~~~~~~~\text{Simplify}~~ \frac{1}{\frac{1}{\csc^2x}}=\csc^2x\\
&=\frac{\cot^2x+1}{\cot^2x-1} ~~~~\because \csc^2x= \cot^2x+1\\
\end{align*}
Since $\sec2x=\frac{\cot^2x+1}{\cot^2x-1}$ therefore, this equation is an identity.
