Answer
$\dfrac{2\tan{x}}{2-\sec^2{x}}=\tan{2x}$
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 $\tan{2x}$.
This means that $\dfrac{2\tan{x}}{2-\sec^2{x}}=\tan{(2x)}$.
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{2\tan{x}}{2-\sec^2{x}}&=\dfrac{2\left(\frac{\sin{x}}{\cos{x}}\right)}{2-\frac{1}{\cos^2{x}}}\\\\
&=\dfrac{\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}}{\cos^2{x}}-\frac{1}{\cos^2{x}}}\\\\
&=\dfrac{\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-1}{\cos^2{x}}}\\\\
&=\frac{2\sin{x}}{\cos{x}} \cdot \frac{\cos^2{x}}{\cos^2{x}-1}\\\\
&=\frac{2\sin{x}}{\cancel{\cos{x}}} \cdot \frac{\cos^\cancel{2}{x}}{\cos^2{x}-1}\\\\
&=2\sin{x} \cdot \frac{\cos{x}}{\cos^2{x}-1}\\\\
&=\frac{2\sin{x}\cos{x}}{\cos^2{x}-1}\\\\
\end{align*}
Recall:
(1) $2\cos^2{A}-1=\cos{(2A)}\\$
(2) $2\sin{A}\cos{A}=\sin{(2A)}$
Use the rules above to obtain:
\begin{align*}
\require{cancel}
\dfrac{2\tan{x}}{2-\sec^2{x}}&=\frac{2\sin{x}\cos{x}}{\cos^2{x}-1}\\\\
&=\dfrac{\sin{2x}}{\cos{2x}}\\\\
&=\tan{2x}
\end{align*}
