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