Answer
$\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{2x}\\\\$
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 $\sin{2x}$.
This means that $\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{(2x)}$.
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{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=\dfrac{4\left(\frac{\sin{x}}{\cos{x}}(\cos^2{x})\right)-2\left(\frac{\sin{x}}{\cos{x}}\right)}{1-\left(\frac{\sin{x}}{\cos{x}}\right)^2}\\\\
&=\dfrac{4\sin{x}\cos{x}-\left(\frac{2\sin{x}}{\cos{x}}\right)}{1-\frac{\sin^2{x}}{\cos^2{x}}}\\\\
&=\dfrac{\frac{(4\sin{x}\cos{x})\cdot \cos{x}}{\cos{x}}-\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}}{\cos^2{x}}-\frac{\sin^2{x}}{\cos^2{x}}}\\\\
&=\dfrac{\frac{4\sin{x}\cos^2{x}}{\cos{x}}-\frac{2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x}}}\\\\
&=\dfrac{\frac{4\sin{x}\cos^2{x}-2\sin{x}}{\cos{x}}}{\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x}}}\\\\
&=\dfrac{4\sin{x}\cos^2{x}-2\sin{x}}{\cos{x}} \cdot \frac{\cos^2{x}}{\cos^2{x}-\sin^2{x}}\\\\
&=\dfrac{4\sin{x}\cos^2{x}-2\sin{x}}{\cancel{\cos{x}}} \cdot \frac{\cos^\cancel{2}{x}}{\cos^2{x}-\sin^2{x}}\\\\
&=\left(4\sin{x}\cos^2{x}-2\sin{x}\right) \cdot \frac{\cos{x}}{\cos^2{x}-\sin^2{x}}\\\\
\end{align*}
Factor out $(2\sin{x})$ to obtain:
\begin{align*}
\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(2\cos^2{x}-1\right) \cdot \frac{\cos{x}}{\cos^2{x}-\sin^2{x}}\\\\
\end{align*}
Recall:
(1) $2\cos^2{A}-1=\cos{(2A)}\\$
(2) $\cos^2{A}-\sin^2{A}=\cos{(2A)}$
Use the rules above to obtain:
\begin{align*}
\require{cancel}
\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(\cos{2x}\right) \cdot \frac{\cos{x}}{\cos{2x}}\\\\
\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\left(\cancel{\cos{2x}}\right) \cdot \frac{\cos{x}}{\cancel{\cos{2x}}}\\\\
\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}&=2\sin{x}\cos{x}\\\\
\end{align*}
Recall:
$\sin{2x}=2\sin{x}\cos{x}$
Thus, using the rule above gives:
$$\dfrac{4\tan{x}\cos^2{x}-2\tan{x}}{1-\tan^2{x}}=\sin{2x}\\\\$$
