Answer
See the explanation below.
Work Step by Step
$\frac{\cot t}{\csc t+1}$
We multiply $\frac{\csc t-1}{\csc t-1}$ with the numerator and denominator of the provided expression:
$\begin{align}
& \frac{\cot t}{\csc t+1}=\frac{\cot t}{\csc t+1}.\frac{\csc t-1}{\csc t-1} \\
& =\frac{\cot t\left( \csc t-1 \right)}{{{\csc }^{2}}t-1}
\end{align}$
Applying the Pythagorean identity of trigonometry ${{\cot }^{2}}t={{\csc }^{2}}t-1$ , which comes out by solving $1+{{\cot }^{2}}t={{\csc }^{2}}t$ , the above expression can be further simplified as:
$\begin{align}
& \frac{\cot t\left( \csc t-1 \right)}{{{\csc }^{2}}t-1}=\frac{\cot t\left( \csc t-1 \right)}{{{\cot }^{2}}t} \\
& =\frac{\csc t-1}{\cot t}
\end{align}$
Thus, the right side of the expression is equal to the left side, which is
$\frac{\csc t-1}{\cot t}=\frac{\cot t}{\csc t+1}$.
