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