Answer
See the explanation below.
Work Step by Step
$\sin 2t-\tan t=\tan t\cos 2t$
Recall the Trigonometric Identities and apply below,
$\begin{align}
& \tan t=\frac{\sin t}{\cos t} \\
& \cos 2t=2{{\cos }^{2}}t-1 \\
& \sin 2t=2\sin t\cos t \\
\end{align}$
Consider the right side of the given expression and apply the above formula,
$\begin{align}
& \tan t\cos 2t=\frac{\sin t}{\cos t}\left( 2{{\cos }^{2}}t-1 \right) \\
& =\frac{\sin t2{{\cos }^{2}}t}{\cos t}-\frac{\sin t}{\cos t} \\
& =2\sin t\cos t-\frac{\sin t}{\cos t} \\
& =\sin 2t-\tan t
\end{align}$
Hence, it is proved that the given identity $\sin 2t-\tan t=\tan t\cos 2t$ holds true.
