[tex]\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)}{1-\sin(t)}[/tex]
Multiply through the second term by [tex]1+\sin(t)[/tex] :
[tex]\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)}{1-\sin(t)}\cdot\dfrac{1+\sin(t)}{1+\sin(t)}[/tex]
[tex]\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)(1+\sin(t))}{1-\sin^2(t)}[/tex]
Recall that [tex]\cos^2(x)+\sin^2(x)=1[/tex] for all x :
[tex]\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{\cos(t)(1+\sin(t))}{\cos^2(t)}[/tex]
If [tex]\cos(t)\neq0[/tex], we can cancel a factor of it in the second fraction.
[tex]\dfrac{1+\sin(t)}{\cos(t)} - \dfrac{1+\sin(t)}{\cos(t)}[/tex]
and this simplifies to
[tex]\dfrac{(1+\sin(t))-(1+\sin(t))}{\cos(t)} = \boxed{0}[/tex]
