Answer
See the explanation below.
Work Step by Step
$\sin t\tan t=\frac{1-{{\cos }^{2}}t}{\cos t}$
Recall Trigonometric Identities,
$\begin{align}
& {{\sin }^{2}}t+{{\cos }^{2}}t=1 \\
& \tan t=\frac{\sin t}{\cos t} \\
\end{align}$
Use the above identities and solve the right side of the given expression,
$\begin{align}
& \frac{1-{{\cos }^{2}}t}{\cos t}=\frac{{{\sin }^{2}}t}{\cos t} \\
& =\sin t\cdot \frac{\sin t}{\cos t} \\
& =\sin t\tan t
\end{align}$
Therefore,
$\sin t\tan t=\frac{1-{{\cos }^{2}}t}{\cos t}$
