Answer
$\dfrac{\cot^2{x}-1}{2\cot{x}}=\cot{2x}$
Work Step by Step
Use a graphing utility to graph the given expression. (Refer to the graph below,)
Notice that the graph is identical with the graph of $\cot{2x}$.
This means that $\dfrac{\cot^2{x}-1}{2\cot{x}}=\cot{(2x)}$.
Work on the left side of the equation above by writing each function using $\sin{x}$ and $\cos{x}$ only:
\begin{align*}
\require{cancel}
\dfrac{\cot^2{x}-1}{2\cot{x}}&=\dfrac{\frac{\cos^2{x}}{\sin^2{x}}-1}{2\left(\frac{\cos{x}}{\sin{x}}\right)}\\\\
&=\dfrac{\frac{\cos^2{x}}{\sin^2{x}}-\frac{\sin^2{x}}{\sin^2{x}}}{\frac{2\cos{x}}{\sin{x}}}\\\\
&=\dfrac{\frac{\cos^2{x}-\sin^2{x}}{\sin^2{x}}}{\frac{2\cos{x}}{\sin{x}}}\\\\
&=\frac{\cos^2{x}-\sin^2{x}}{\sin^2{x}}\cdot \frac{\sin{x}}{2\cos{x}}\\\\
&=\frac{\cos^2{x}-\sin^2{x}}{\sin^\cancel{2}{x}}\cdot \frac{\cancel{\sin{x}}}{2\cos{x}}\\\\
&=\frac{\cos^2{x}-\sin^2{x}}{\sin{x}}\cdot \frac{1}{2\cos{x}}\\\\
&=\frac{\cos^2{x}-\sin^2{x}}{2\sin{x}\cos{x}}\\\\
\end{align*}
Recall:
(1) $\cos^2{A}-\sin^2{A}=\cos{(2A)}\\$
(2) $2\sin{A}\cos{A}=\sin{(2A)}$
Use the rules above to obtain:
\begin{align*}
\require{cancel}
\dfrac{\cot^2{x}-1}{2\cot{x}}&=\frac{\cos^2{x}-\sin^2{x}}{2\sin{x}\cos{x}}\\\\
&=\frac{\cos{2x}}{\sin{2x}}\\\\
&=\cot{2x}
\end{align*}
