Answer
See the explanation below.
Work Step by Step
$\frac{1+\cos t}{1-\cos t}$
We multiply $\frac{1+\cos t}{1+\cos t}$ to the provided expression:
$\begin{align}
& \frac{1+\cos t}{1-\cos t}=\frac{1+\cos t}{1-\cos t}.\frac{1+\cos t}{1+\cos t} \\
& =\frac{{{\left( 1+\cos t \right)}^{2}}}{1-{{\cos }^{2}}t}
\end{align}$
Applying the Pythagorean identity of trigonometry ${{\sin }^{2}}t=1-{{\cos }^{2}}t$ , which comes out by solving ${{\sin }^{2}}t+{{\cos }^{2}}t=1$ , the above expression can be further simplified as:
$\frac{{{\left( 1+\cos t \right)}^{2}}}{1-{{\cos }^{2}}t}=\frac{{{\left( 1-\cos t \right)}^{2}}}{{{\sin }^{2}}t}$
Now, consider the right side of the given expression:
${{\left( \csc t-\cot t \right)}^{2}}$
By using the reciprocal identity $\csc t=\frac{1}{\sin t}$ , and the quotient identity of trigonometry $\cot t=\frac{\cos t}{\sin t}$ , now, the above expression can be further simplified as:
$\begin{align}
& {{\left( \csc t-\cot t \right)}^{2}}={{\left( \frac{1}{\sin t}-\frac{\cos t}{\sin t} \right)}^{2}} \\
& ={{\left( \frac{1-\cos t}{\sin t} \right)}^{2}} \\
& =\frac{{{\left( 1-\cos t \right)}^{2}}}{{{\sin }^{2}}t}
\end{align}$
Now, the identity is verified because both sides are equal to $\frac{{{\left( 1-\cos t \right)}^{2}}}{{{\sin }^{2}}t}$.
Thus, the left side of the expression is equal to the right side, which is
$\frac{1+\cos t}{1-\cos t}={{\left( \csc t-\cot t \right)}^{2}}$.
