Answer
The identity is true.
For proof, please see "step-by-step".
Work Step by Step
Recognize a sum of cubes in the numerator.
$x^{3}+y^{y}=(x+y)(x^{2}-xy+y^{2})$
$LHS=\displaystyle \frac{1+\cot^{3}t}{1+\cot t}=$
... substituting $x=1, y=\cot t$
$\displaystyle \frac{x^{3}+y^{3}}{x+y}=\frac{(x+y)(x^{2}-xy+y^{2})}{(x+y)}$
... reduce
$=x^{2}-xy+y^{2}$
...back substitute
$=1^{2}-\cot t+\cot^{2}t$
$...$ apply Pyth. identity: $ 1+\cot^{2}\theta=\csc^{2}\theta$
$=\csc^{2}\theta-\cot t=RHS$,
proving the identity.