Answer
See the explanation below.
Work Step by Step
To verify the given identity,
$\csc x-\csc x{{\sin }^{2}}x=\sin x$
Recall Trigonometric Identities,
$\begin{align}
& \csc x=\frac{1}{\sin x} \\
& {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
\end{align}$
Use the above identities and solve the left side of the given expression,
$\begin{align}
& \csc x-\csc x{{\sin }^{2}}x=\csc x\left( 1-{{\cos }^{2}}x \right) \\
& =\left( \frac{1}{\sin x} \right)\left( {{\sin }^{2}}x \right) \\
& =\sin x
\end{align}$
Hence, it is proved that the given identity holds true.
