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