Answer
The right side is equal to $\cos 2x$.
Work Step by Step
By using the trigonometric identity, $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{sinx}{\cos x}$.
$\begin{align}
& \frac{\cot x-\tan x}{\cot x+\tan x}=\frac{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}} \\
& =\frac{\frac{\cos x}{\sin x}\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}\frac{\sin x}{\sin x}}{\frac{\cos x}{\sin x}\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}\frac{\sin x}{\sin x}} \\
& =\frac{\frac{{{\cos }^{2}}x}{\sin x\cos x}-\frac{{{\sin }^{2}}x}{\sin x\cos x}}{\frac{{{\cos }^{2}}x}{\sin x\cos x}-\frac{{{\sin }^{2}}x}{\sin x\cos x}} \\
& =\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x}
\end{align}$
Now, apply trigonometric identities,
${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
Simplifying it further,
$\begin{align}
& \frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x+{{\sin }^{2}}x}=\frac{\cos 2x}{1} \\
& =\cos 2x
\end{align}$
Thus, the right side of the equation is equal to $\cos 2x$.
Thus, it is proved that the left and right sides are equal.
