Answer
The expression on the left side is equal to the expression on the right side.
Work Step by Step
The given expression on the left side $\frac{1-\cos t}{1+\cos t}$ can be simplified by rationalizing it.
$\begin{align}
& \frac{1-\cos t}{1+\cos t}=\frac{1-\cos t}{1+\cos t}.\frac{1-\cos t}{1-\cos t} \\
& =\frac{{{\left( 1-\cos t \right)}^{2}}}{1-{{\cos }^{2}}t} \\
& =\frac{{{\left( 1-\cos t \right)}^{2}}}{{{\sin }^{2}}t} \\
& ={{\left( \frac{1-\cos t}{\sin t} \right)}^{2}}
\end{align}$
Now, the expression can be simplified by applying the reciprocal identity $\frac{1}{\sin t}=\csc t$ and quotient identity $\cot t=\frac{\cos t}{\sin t}$
$\begin{align}
& {{\left( \frac{1-\cos t}{\sin t} \right)}^{2}}={{\left( \frac{1}{\sin t}-\frac{\cos t}{\sin t} \right)}^{2}} \\
& ={{\left( \csc t-\cot \right)}^{2}}
\end{align}$
Hence, the expression on the left side is equal to the expression on the right side.
