$\displaystyle
\frac{1+cot^3 t}{1+cot t}=csc^2 t - cot t
$
Hello, john doe!
Mr. F is absolutely correct . . .
Factor that sum-of-cubes . . .$\displaystyle \frac{1+\cot^3\!t}{1+\cot t}\:=\:\csc^2\!t - \cot t$
$\displaystyle \text{We have: }\;\frac{(1+\cot t)(1 - \cot t + \cot^2\!t)}{1 + \cot t} \;\;=\;\;\underbrace{1 + \cot^2\!t}_{\text{This is }\csc^2\!t} - \cot t \;\;=\;\;\csc^2\!t - \cot t$