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