Answer
See the explanation below.
Work Step by Step
$\frac{{{\cot }^{2}}t}{\csc t}=\csc t-\sin t$
Recall Trigonometric Identities,
$\begin{align}
& 1+{{\cot }^{2}}t={{\csc }^{2}}t \\
& \csc t=\frac{1}{\sin t} \\
\end{align}$
Consider the left side of the given expression,
$\begin{align}
& \frac{{{\cot }^{2}}t}{\csc t}=\frac{{{\csc }^{2}}t-1}{\csc t} \\
& =\csc t-\frac{1}{\csc t} \\
& =\csc t-\sin t
\end{align}$
Therefore,
$\frac{{{\cot }^{2}}t}{\csc t}=\csc t-\sin t$
